BE Civil Engineering (IOE, TU) Sanitary Engineering (IOE, CE 654) Question Paper 2080 Nepal

(a) Flow estimation

Average wastewater flow from population

$$Q_{avg} = \frac{P \times q \times 0.80}{1000} = \frac{24000 \times 135 \times 0.80}{1000} = 2592\ \text{m}^3/\text{day}$$

Add infiltration (12% of average DWF from sewage):

$$Q_{inf} = 0.12 \times 2592 = 311.04\ \text{m}^3/\text{day}$$ $$Q_{DWF} = 2592 + 311.04 = 2903.04\ \text{m}^3/\text{day}$$

Convert to $\text{m}^3/\text{s}$:

$$Q_{DWF} = \frac{2903.04}{86400} = 0.0336\ \text{m}^3/\text{s}$$

Peak factor (Harmon's formula): with population in thousands $p = 24$,

$$P = 1 + \frac{14}{4 + \sqrt{p}} = 1 + \frac{14}{4 + \sqrt{24}} = 1 + \frac{14}{4 + 4.899} = 1 + 1.573 = 2.573$$

Peak flow:

$$Q_{peak} = 2.573 \times 0.0336 = 0.0865\ \text{m}^3/\text{s}$$

Minimum flow (taken as $\tfrac{1}{3}$ of average DWF, common IOE practice):

$$Q_{min} = \frac{1}{3} \times 0.0336 = 0.0112\ \text{m}^3/\text{s}$$

(b) Sewer design for peak flow running full

Manning's equation for a circular pipe flowing full (diameter $D$):

$$Q = \frac{1}{n} A R^{2/3} S^{1/2}, \quad A = \frac{\pi D^2}{4}, \quad R = \frac{D}{4}$$

Adopt a self-cleansing slope. Try $D = 0.30\ \text{m}$ and solve for the slope giving $Q = 0.0865\ \text{m}^3/\text{s}$.

$$A = \frac{\pi (0.30)^2}{4} = 0.0707\ \text{m}^2, \quad R = \frac{0.30}{4} = 0.075\ \text{m}$$ $$R^{2/3} = 0.075^{0.667} = 0.1782$$

Required slope:

$$S^{1/2} = \frac{Q\,n}{A R^{2/3}} = \frac{0.0865 \times 0.013}{0.0707 \times 0.1782} = \frac{0.0011245}{0.012599} = 0.08925$$ $$S = 0.007965 \approx \frac{1}{125}$$

Velocity at full (peak) flow:

$$V_{full} = \frac{Q}{A} = \frac{0.0865}{0.0707} = 1.224\ \text{m/s} \;>\; 0.6\ \text{m/s} \;\checkmark$$

So $D = 300\ \text{mm}$ at slope $\approx 1\text{ in }125$ carries the peak flow just full at $1.22\,\text{m/s}$.

Check velocity at minimum flow

Proportional depth: $\dfrac{q}{Q_{full}} = \dfrac{0.0112}{0.0865} = 0.1295$.

From the hydraulic-elements (proportionate) chart for a circular sewer, $q/Q \approx 0.13$ corresponds to $d/D \approx 0.25$ and $v/V \approx 0.70$.

$$v_{min} \approx 0.70 \times 1.224 = 0.857\ \text{m/s} \;>\; 0.6\ \text{m/s} \;\checkmark$$

Result: A 300 mm diameter sewer laid at a grade of about 1 in 125 satisfies self-cleansing velocity at both peak and minimum flow.

Given

• $Q = 6000\ \text{m}^3/\text{day}$, $S_0 = 220\ \text{mg/L}$, $S = 20\ \text{mg/L}$
• $X =$ MLSS $= 3000\ \text{mg/L}$, $\theta_c = 8\ \text{d}$, $Y = 0.6$, $k_d = 0.06\ \text{d}^{-1}$

BOD removed: $S_0 - S = 220 - 20 = 200\ \text{mg/L}$.

(a) Aeration tank volume

Standard ASP design relation:

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

Numerator (work in consistent mg/L · m³/d, the L cancels):

$$\theta_c Q Y (S_0-S) = 8 \times 6000 \times 0.6 \times 200 = 5{,}760{,}000$$

Denominator:

$$X(1 + k_d\theta_c) = 3000 \times (1 + 0.06\times 8) = 3000 \times 1.48 = 4440$$ $$V = \frac{5{,}760{,}000}{4440} = 1297.3\ \text{m}^3 \approx \mathbf{1300\ m^3}$$

Hydraulic retention time:

$$t = \frac{V}{Q} = \frac{1297.3}{6000} = 0.2162\ \text{day} = 5.19\ \text{hours}$$

(b) F/M ratio

$$\frac{F}{M} = \frac{Q\,S_0}{V\,X} = \frac{6000 \times 220}{1297.3 \times 3000} = \frac{1{,}320{,}000}{3{,}891{,}900} = 0.339\ \text{day}^{-1}$$

(Within the conventional range $0.2$–$0.5\ \text{d}^{-1}$.)

(c) Waste activated sludge

Observed yield:

$$Y_{obs} = \frac{Y}{1 + k_d \theta_c} = \frac{0.6}{1.48} = 0.4054$$

Mass of sludge wasted per day:

$$P_x = Y_{obs}\,Q\,(S_0 - S) = 0.4054 \times 6000 \times 200\ \text{g/day}$$ $$= 0.4054 \times 6000 \times 200 \times 10^{-3}\ \text{kg/day} = 486.5\ \text{kg/day}$$

Waste activated sludge $\approx \mathbf{486.5\ kg\ VSS/day}$.

(Check via sludge age: $P_x = \dfrac{X\,V}{\theta_c} = \dfrac{3000\times 1297.3 \times 10^{-3}}{8} = 486.5\ \text{kg/day}$ — consistent.)

(a) BOD vs COD

Biodegradability index: A high $\text{BOD}_5/\text{COD}$ ratio (typically $> 0.5$) means most of the organic load is biodegradable, so biological treatment is suitable. A low ratio ($< 0.3$) indicates a large non-biodegradable/toxic fraction, favouring physico-chemical treatment.

(b) Ultimate BOD and remaining BOD

First-order BOD model (base $e$):

$$\text{BOD}_t = L_0\left(1 - e^{-kt}\right)$$

At $t = 5$ d, $\text{BOD}_5 = 180\ \text{mg/L}$, $k = 0.23\ \text{d}^{-1}$:

$$e^{-kt} = e^{-0.23 \times 5} = e^{-1.15} = 0.3166$$ $$1 - e^{-kt} = 1 - 0.3166 = 0.6834$$ $$L_0 = \frac{\text{BOD}_5}{0.6834} = \frac{180}{0.6834} = 263.4\ \text{mg/L}$$

Ultimate first-stage BOD $L_0 \approx \mathbf{263.4\ mg/L}$.

BOD remaining (unexerted) after 5 days:

$$L_t = L_0\,e^{-kt} = 263.4 \times 0.3166 = 83.4\ \text{mg/L}$$

(Check: $263.4 - 83.4 = 180\ \text{mg/L} = \text{BOD}_5$ ✓)

BOD remaining after 5 days $\approx \mathbf{83.4\ mg/L}$.

(a) Functional elements of MSWM

1. Waste generation – activities that identify materials as no longer useful.
2. On-site handling, storage & processing – sorting and storing waste at source.
3. Collection – gathering waste and hauling to transfer/processing/disposal points.
4. Transfer & transport – moving waste from small collection vehicles to large ones / transfer stations for economical long-haul.
5. Processing & recovery – separation, recycling, composting, energy recovery, volume reduction.
6. Disposal – final placement, typically in an engineered sanitary landfill.

These elements are managed in an integrated manner following the waste hierarchy: reduce → reuse → recycle → recover → dispose.

(b) Quantities and truck trips

Daily mass of waste:

$$M = 90{,}000 \times 0.45 = 40{,}500\ \text{kg/day} = \mathbf{40.5\ tonnes/day}$$

Daily volume (as-collected):

$$V = \frac{M}{\rho} = \frac{40{,}500}{300} = 135\ \text{m}^3/\text{day}$$

Number of truck trips:

$$N = \frac{V}{\text{truck capacity}} = \frac{135}{8} = 16.875$$

Round up to the next whole trip:

$$N = \mathbf{17\ truck\ trips\ per\ day}$$

(If a fleet operates and each truck makes, say, 3 trips/day, then $\lceil 17/3 \rceil = 6$ trucks would be required.)

Given

• $N = 400$ students, $q = 25\ \text{L/student/day}$
• Detention time $T = 24\ \text{h} = 1\ \text{day}$
• Sludge accumulation $= 30\ \text{L/student/year}$, desludging every 2 years
• Liquid depth $= 1.5\ \text{m}$, free board $= 0.3\ \text{m}$, $L:B = 3:1$

Step 1 – Daily wastewater flow

$$Q = N \times q = 400 \times 25 = 10{,}000\ \text{L/day} = 10\ \text{m}^3/\text{day}$$

Step 2 – Liquid (sedimentation) volume from detention time

$$V_l = Q \times T = 10 \times 1 = 10\ \text{m}^3$$

Step 3 – Sludge storage volume

$$V_s = \frac{N \times \text{accumulation} \times \text{interval}}{1000} = \frac{400 \times 30 \times 2}{1000} = 24\ \text{m}^3$$

Step 4 – Total required volume

$$V = V_l + V_s = 10 + 24 = 34\ \text{m}^3$$

Step 5 – Plan area and dimensions

Using liquid depth $= 1.5\ \text{m}$:

$$A = \frac{V}{\text{depth}} = \frac{34}{1.5} = 22.67\ \text{m}^2$$

With $L = 3B$:

$$L \times B = 3B^2 = 22.67 \Rightarrow B^2 = 7.556 \Rightarrow B = 2.75\ \text{m}$$ $$L = 3 \times 2.75 = 8.25\ \text{m}$$

Provide $L = 8.25\ \text{m}$, $B = 2.75\ \text{m}$.

Step 6 – Total depth

$$\text{Total depth} = \text{liquid depth} + \text{free board} = 1.5 + 0.3 = 1.8\ \text{m}$$

Final septic tank size: $\mathbf{8.25\ m \times 2.75\ m \times 1.8\ m}$ (with 1.5 m liquid depth + 0.3 m free board).

Sewer appurtenances

Appurtenances are accessory structures provided along a sewer line for its proper functioning, inspection, cleaning, ventilation and flow management. Common types: manholes, drop manholes, lamp holes, clean-outs, inlets/catch basins, flushing tanks, grease & oil traps, inverted siphons, and storm overflows.

Manhole

A masonry/RCC chamber giving access to the sewer for inspection, cleaning and maintenance. Provided at every change of direction, gradient, size, at junctions, and on straight runs at regular spacing (e.g. 30 m for small sewers up to ~90 m for large ones). It also serves for ventilation.

   ground level   ___[cover & frame]___
                  |   neck/access     |
                  |   working chamber |
   benching ----> \__channel(invert)__/  flow ->

Drop manhole

Used when a branch/incoming sewer joins the main at a level much higher than the main (steep terrain). Instead of letting sewage cascade and erode the invert, a vertical drop pipe brings the flow down to the channel level smoothly, reducing turbulence and splashing.

  high inlet --->| drop pipe |
                 |   ||      |
  main sewer ____|___\/______|___ ->

Inverted siphon

A depressed portion of sewer that runs full under pressure to pass below an obstruction (stream, road, valley, utility). Flow is driven by the head difference between inlet and outlet. Designed with self-cleansing velocity (≥ 0.9–1.0 m/s) to prevent silting, often with multiple barrels for varying flows and inlet/outlet chambers for cleaning.

  inlet ___                         ___ outlet
         \                         /
          \___ obstruction over __/  (pipe runs full)

Given

$Q = 4000\ \text{m}^3/\text{day}$, SOR $= 32\ \text{m}^3/\text{m}^2/\text{day}$, detention $t = 2.5\ \text{h}$.

Surface area

$$A = \frac{Q}{\text{SOR}} = \frac{4000}{32} = 125\ \text{m}^2$$

Diameter

$$A = \frac{\pi D^2}{4} \Rightarrow D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 125}{\pi}} = \sqrt{159.15} = 12.62\ \text{m}$$

Provide $D \approx 12.6\ \text{m}$.

Depth from detention time

Volume required:

$$V = Q \times t = 4000 \times \frac{2.5}{24} = 416.7\ \text{m}^3$$ $$\text{Depth} = \frac{V}{A} = \frac{416.7}{125} = 3.33\ \text{m}$$

Provide liquid depth $\approx 3.35\ \text{m}$ (add ~0.3 m free board → total ~3.65 m).

Weir loading (peripheral weir)

Weir length = circumference:

$$L_w = \pi D = \pi \times 12.62 = 39.65\ \text{m}$$ $$\text{Weir loading} = \frac{Q}{L_w} = \frac{4000}{39.65} = 100.9\ \text{m}^3/\text{m}/\text{day}$$

Since $100.9 < 180\ \text{m}^3/\text{m}/\text{day}$, the peripheral weir loading is acceptable — a single peripheral effluent weir suffices.

Summary: $A = 125\ \text{m}^2$, $D \approx 12.6\ \text{m}$, depth $\approx 3.35\ \text{m}$, weir loading $\approx 101\ \text{m}^3/\text{m}/\text{day}$ (OK).

(a) Sludge digestion

Sludge digestion is the biological stabilization of organic sludge from treatment units, reducing its volume, pathogen content and putrescibility so it can be safely dewatered and disposed of.

(b) Three phases of anaerobic digestion

1. Hydrolysis – complex organics (carbohydrates, proteins, fats) are broken down by enzymes into soluble simpler compounds (sugars, amino acids, fatty acids).
2. Acidogenesis / Acetogenesis – acid-forming bacteria convert the products into volatile fatty acids, $H_2$ and $CO_2$, and then to acetic acid.
3. Methanogenesis – methanogenic archaea convert acetic acid, $H_2$ and $CO_2$ into methane ($CH_4$) and $CO_2$.

Main useful product: biogas (≈ 60–70% methane), usable as fuel.

Two operating parameters to control:

• Temperature (mesophilic ≈ 35°C or thermophilic ≈ 55°C; must be steady).
• pH (maintain ≈ 6.8–7.2; methanogens are sensitive to acidity). (Others: HRT/sludge age, mixing, absence of toxins.)

(a) Tertiary / advanced treatment

Tertiary treatment is the treatment applied after secondary (biological) treatment to remove residual pollutants that conventional primary and secondary stages cannot — such as nutrients (N, P), suspended solids, dissolved solids, refractory organics, and pathogens.

Why needed:

• To meet stringent effluent/discharge standards for sensitive receiving waters.
• To prevent eutrophication caused by nitrogen and phosphorus.
• To enable water reuse (irrigation, industrial, recharge).
• To protect public health by removing pathogens.

(b) Two tertiary processes

1. Nitrogen removal (Nitrification–Denitrification):

• Nitrification (aerobic, autotrophic bacteria): $NH_4^+ \rightarrow NO_2^- \rightarrow NO_3^-$.
• Denitrification (anoxic, heterotrophic bacteria): $NO_3^- \rightarrow N_2$ gas (released to atmosphere).
• Purpose: removes ammonia (toxic, oxygen-demanding) and nitrate (eutrophication, health) from effluent.

2. Disinfection (e.g., chlorination / UV / ozone):

• Destroys or inactivates pathogenic micro-organisms in the treated effluent.
• Chlorination is common and cheap (needs contact time, may leave residual; risk of DBPs); UV leaves no residual and forms no by-products.
• Purpose: protect public health and downstream water uses by reducing bacteria/virus counts.

(Alternative — Phosphorus removal: by chemical precipitation with alum/lime/ferric salts or by enhanced biological phosphorus removal (EBPR); purpose: control eutrophication.)

Given

$Q = 2500\ \text{m}^3/\text{day}$, $\text{BOD}_5 = 180\ \text{mg/L}$, organic loading $= 0.20\ \text{kg/m}^3/\text{d}$, hydraulic loading $= 2.0\ \text{m}^3/\text{m}^2/\text{d}$.

BOD load applied

$$\text{BOD load} = Q \times \text{conc} = 2500\ \tfrac{m^3}{d} \times 180\ \tfrac{g}{m^3} = 450{,}000\ \text{g/day} = 450\ \text{kg/day}$$

Filter volume (from organic loading)

$$V = \frac{\text{BOD load}}{\text{organic loading}} = \frac{450}{0.20} = 2250\ \text{m}^3$$

Surface area (from hydraulic loading)

$$A = \frac{Q}{\text{hydraulic loading}} = \frac{2500}{2.0} = 1250\ \text{m}^2$$

Depth

$$\text{Depth} = \frac{V}{A} = \frac{2250}{1250} = 1.8\ \text{m}$$

This depth (~1.8 m) is typical for a low-rate trickling filter (usual range 1.5–3.0 m).

Result: Volume $= 2250\ \text{m}^3$, Surface area $= 1250\ \text{m}^2$, Depth $= 1.8\ \text{m}$.

(If a single circular filter is used: $D = \sqrt{4A/\pi} = \sqrt{4\times1250/\pi} = 39.9\ \text{m}$; in practice two or more units would be provided.)

(a) Sanitary landfill

A sanitary landfill is an engineered method of disposing of solid waste on land by spreading it in thin layers, compacting it to the smallest practical volume, and covering it with soil daily, in a manner that protects the environment and public health.

Essential components:

• Liner system (clay/HDPE geomembrane) to prevent groundwater contamination.
• Leachate collection & treatment system.
• Landfill gas (methane) collection/venting system.
• Daily, intermediate and final cover.
• Surface-water drainage and monitoring wells.

(b) Land area required per year

Daily volume in place:

$$V_{day} = \frac{\text{mass}}{\text{density}} = \frac{50{,}000\ \text{kg/day}}{600\ \text{kg/m}^3} = 83.33\ \text{m}^3/\text{day}$$

Annual volume:

$$V_{year} = 83.33 \times 365 = 30{,}417\ \text{m}^3/\text{year}$$

Land area (depth 5 m):

$$A = \frac{V_{year}}{\text{depth}} = \frac{30{,}417}{5} = 6083.3\ \text{m}^2/\text{year}$$

Land area required $\approx \mathbf{6083\ m^2/year}$ (about 0.61 hectare/year) for the waste alone; allowing for cover soil the actual requirement would be ~15–25% higher.