BE Civil Engineering (IOE, TU) Sanitary Engineering (IOE, CE 654) Question Paper 2076 Nepal

(a) Average dry weather flow (DWF)

Water demand $= 45{,}000 \times 135 = 6{,}075{,}000\ \text{L/day} = 6075\ \text{m}^3/\text{day}$.

Wastewater fraction $= 80\%$:

$$Q_{avg} = 0.80 \times 6075 = 4860\ \text{m}^3/\text{day}$$

Converting to L/s:

$$Q_{avg} = \frac{4860 \times 1000}{86400} = 56.25\ \text{L/s}$$

Average DWF = 4860 m³/day = 56.25 L/s

(b) Peak factor (Harmon's formula)

Population in thousands $p = 45$, so $\sqrt{p} = \sqrt{45} = 6.708$.

$$P_f = 1 + \frac{14}{4 + 6.708} = 1 + \frac{14}{10.708} = 1 + 1.3075 = 2.308$$

Peak sewage flow:

$$Q_{peak} = 2.308 \times 56.25 = 129.8\ \text{L/s}$$

Peak factor = 2.31; Peak sanitary flow = 129.8 L/s

(c) Total maximum design flow

Infiltration $= 5000\ \text{L/ha/day} \times 210\ \text{ha} = 1{,}050{,}000\ \text{L/day}$

$$= \frac{1{,}050{,}000}{86400} = 12.15\ \text{L/s}$$

Total maximum design flow:

$$Q_{design} = Q_{peak} + Q_{infiltration} = 129.8 + 12.15 = 141.95\ \text{L/s}$$

Total maximum design flow ≈ 142 L/s

(a) Sewer flowing full

Diameter $D = 0.40\ \text{m}$, slope $S = 1/500 = 0.002$.

Area (full): $A = \frac{\pi D^2}{4} = \frac{\pi (0.40)^2}{4} = 0.12566\ \text{m}^2$

Hydraulic radius (full): $R = D/4 = 0.40/4 = 0.10\ \text{m}$

Velocity:

$$V_{full} = \frac{1}{0.013}(0.10)^{2/3}(0.002)^{1/2}$$

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

$$V_{full} = \frac{1}{0.013} \times 0.21544 \times 0.044721 = \frac{0.009635}{0.013} = 0.741\ \text{m/s}$$

Discharge:

$$Q_{full} = A \times V_{full} = 0.12566 \times 0.741 = 0.0931\ \text{m}^3/\text{s} = 93.1\ \text{L/s}$$

Q_full = 93.1 L/s, V_full = 0.741 m/s

(b) Sewer flowing at depth d = 0.5 D (half full)

For $d/D = 0.5$, the hydraulic elements (proportional ratios) are:

• $A/A_{full} = 0.5$ (half the area)
• $R/R_{full} = 1.0$ (hydraulic radius at half depth equals that at full)
• $V/V_{full} = 1.0$
• $Q/Q_{full} = 0.5$

This follows because at half depth the wetted perimeter is half of full and area is half of full, so $R = A/P$ is identical to the full-flow value.

Velocity at half full:

$$V_{0.5} = 1.0 \times 0.741 = 0.741\ \text{m/s}$$

Discharge at half full:

$$Q_{0.5} = 0.5 \times 93.1 = 46.6\ \text{L/s}$$

Comment: $V_{0.5} = 0.741\ \text{m/s} > 0.6\ \text{m/s}$, so the self-cleansing velocity IS satisfied at half-full flow. The sewer will resist deposition of suspended solids.

Q_(0.5D) = 46.6 L/s, V_(0.5D) = 0.741 m/s — self-cleansing achieved.

Given: $Q = 6\ \text{MLD} = 6000\ \text{m}^3/\text{day}$.

(a) Surface area and plan dimensions

Surface area from SOR:

$$A = \frac{Q}{SOR} = \frac{6000}{30} = 200\ \text{m}^2$$

With $L = 4B$: $A = L \times B = 4B \times B = 4B^2 = 200$

$$B^2 = 50 \Rightarrow B = 7.07\ \text{m}, \quad L = 4 \times 7.07 = 28.28\ \text{m}$$

Adopt B = 7.1 m, L = 28.4 m (area ≈ 201.6 m²).

(b) Depth from detention time

Volume required:

$$V = Q \times t = 6000\ \text{m}^3/\text{day} \times \frac{2.0}{24}\ \text{day} = 6000 \times 0.08333 = 500\ \text{m}^3$$

Depth:

$$H = \frac{V}{A} = \frac{500}{200} = 2.5\ \text{m}$$

Provide a free board of 0.5 m → total tank depth = 3.0 m; liquid depth = 2.5 m.

(c) Weir loading check

Effluent weir along the width $B = 7.07\ \text{m}$:

$$\text{Weir loading} = \frac{Q}{B} = \frac{6000}{7.07} = 848.7\ \text{m}^3/\text{m/day}$$

This exceeds the limit of 180 m³/m/day. To satisfy the criterion, increase weir length. Required weir length:

$$L_w = \frac{6000}{180} = 33.3\ \text{m}$$

Provide multiple weir troughs / launders (e.g., a finger-weir or suspended weir trough giving ≈ 34 m of crest, achieved by a multi-pass launder across the tank width).

Final design: L = 28.4 m, B = 7.1 m, liquid depth = 2.5 m (3.0 m total), surface area ≈ 200 m², with effluent launders giving ≥ 33.3 m weir crest to meet the weir-loading limit.

Given: $Q = 4\ \text{MLD} = 4000\ \text{m}^3/\text{day}$, $S_0 = 220$, $S = 20\ \text{mg/L}$.

(a) Aeration tank volume

The complete-mix design equation:

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

Numerator: $\theta_c Q Y (S_0-S) = 8 \times 4000 \times 0.5 \times (220-20) = 8 \times 4000 \times 0.5 \times 200 = 3{,}200{,}000$ (units g/m³·m³ → using mg/L = g/m³).

Denominator: $X(1 + k_d\theta_c) = 3000 \times (1 + 0.06 \times 8) = 3000 \times 1.48 = 4440$

$$V = \frac{3{,}200{,}000}{4440} = 720.7\ \text{m}^3$$

Hydraulic retention time:

$$\theta = \frac{V}{Q} = \frac{720.7}{4000} = 0.1802\ \text{day} = 4.32\ \text{hours}$$

V = 721 m³, HRT = 4.3 h

(b) Waste sludge production

Observed yield:

$$Y_{obs} = \frac{Y}{1 + k_d \theta_c} = \frac{0.5}{1.48} = 0.3378$$

Mass of sludge wasted per day:

$$P_x = Y_{obs}\, Q (S_0 - S) = 0.3378 \times 4000 \times 200 \times 10^{-3}\ \text{kg/day}$$ $$P_x = 0.3378 \times 4000 \times 200 / 1000 = 270.3\ \text{kg/day}$$

(Here $Q(S_0-S)$ in g/day = $4000 \times 200 = 800{,}000\ \text{g/day}$; $\times 0.3378 = 270{,}300\ \text{g/day} = 270.3\ \text{kg/day}$.)

Volume of waste sludge at $X_r = 10{,}000\ \text{mg/L} = 10\ \text{kg/m}^3$:

$$Q_w = \frac{P_x}{X_r} = \frac{270.3}{10} = 27.0\ \text{m}^3/\text{day}$$

Sludge mass = 270.3 kg/day; sludge volume ≈ 27.0 m³/day

(c) F/M ratio

$$F/M = \frac{Q\, S_0}{V\, X} = \frac{4000 \times 220}{720.7 \times 3000} = \frac{880{,}000}{2{,}162{,}100} = 0.407\ \text{d}^{-1}$$

F/M ≈ 0.41 kg BOD / kg MLSS·day — within the typical conventional ASP range (0.2–0.5).

(a) Daily generation and loose volume

Mass per day:

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

Loose volume:

$$V_{loose} = \frac{54{,}000}{180} = 300\ \text{m}^3/\text{day}$$

Generation = 54 tonnes/day; loose volume = 300 m³/day

(b) Number of collection trucks

Compaction ratio 2.5 means the in-truck (compacted) density is 2.5× loose; therefore the effective volume occupied per day after compaction:

$$V_{compacted} = \frac{300}{2.5} = 120\ \text{m}^3/\text{day}$$

Volume hauled per truck per day = capacity × trips = $8 \times 3 = 24\ \text{m}^3/\text{truck/day}$.

Number of trucks:

$$N = \frac{120}{24} = 5\ \text{trucks}$$

5 trucks required.

(c) ISWM hierarchy and the 3R principle

The ISWM hierarchy ranks waste-handling options from most to least preferred:

1. Source reduction / waste prevention — avoid generating waste (most preferred).
2. Reuse — use items again without reprocessing.
3. Recycling & composting — recover materials and organics.
4. Energy recovery — incineration with energy capture, waste-to-energy, biogas.
5. Treatment & disposal — sanitary landfilling of residuals (least preferred).

The 3R principle (Reduce, Reuse, Recycle) occupies the top three tiers of the hierarchy, emphasising minimisation of waste at source before any treatment or disposal. Effective ISWM also integrates segregation at source, transfer stations, public participation, institutional/financial arrangements, and regulatory enforcement.

(a) Ultimate BOD $L_0$

$$BOD_5 = L_0 (1 - e^{-k \cdot 5})$$

$e^{-0.23 \times 5} = e^{-1.15} = 0.31664$

$$180 = L_0 (1 - 0.31664) = L_0 \times 0.68336$$ $$L_0 = \frac{180}{0.68336} = 263.4\ \text{mg/L}$$

$L_0$ = 263.4 mg/L

(b) BOD exerted after 3 days

$e^{-0.23 \times 3} = e^{-0.69} = 0.50158$

$$BOD_3 = 263.4 \times (1 - 0.50158) = 263.4 \times 0.49842 = 131.3\ \text{mg/L}$$

$BOD_3$ = 131.3 mg/L

(a) Common sewer appurtenances and their functions

(Any four with correct function earn full marks.)

(b) Drop manhole

A drop manhole is a special manhole used where a branch (incoming) sewer joins a main sewer at a substantially higher elevation — typically when the difference in invert levels is greater than about 0.6 m.

Sketch (described):

        incoming sewer (high invert)
            |
   =========+=========   <- manhole top / cover
   ||       |       ||
   ||   vertical    ||
   ||   drop pipe   ||
   ||      ||       ||
   ||      ||       ||
   ||      \/  ===> outgoing main sewer (low invert)
   ==================
        manhole base / channel

Purpose / working: Instead of allowing the high-level sewage to cascade freely down the manhole face (which causes splashing, erosion of the benching, and release of foul gases / turbulence), the flow is led down through a vertical drop pipe (down-take pipe) built outside or inside the manhole chamber. The sewage is delivered gently to the invert of the lower main sewer. This:

• avoids erosion and damage to the manhole bottom,
• reduces splashing and odour nuisance,
• prevents the sewer worker from being exposed to falling sewage during maintenance,
• helps dissipate energy where there is a steep drop in ground level.

Condition: Provided when the incoming sewer invert is appreciably higher (typically > 0.6 m) than the outgoing sewer invert, common in hilly/steep terrain like much of Nepal.

Step 1 — Volume for detention (liquid/sewage flow component)

Daily sewage flow:

$$Q = 8 \times 90 = 720\ \text{L/day} = 0.72\ \text{m}^3/\text{day}$$

Detention volume (for 2 days):

$$V_1 = Q \times t = 0.72 \times 2 = 1.44\ \text{m}^3$$

Step 2 — Volume for sludge & scum storage

$$V_2 = 0.04\ \text{m}^3/\text{capita/year} \times 8\ \text{persons} \times 3\ \text{years} = 0.96\ \text{m}^3$$

Step 3 — Total effective volume

$$V = V_1 + V_2 = 1.44 + 0.96 = 2.40\ \text{m}^3$$

Total effective (liquid) capacity required = 2.40 m³

A typical proportioning: provide a liquid depth of about 1.2–1.8 m, length-to-width ratio of 2:1 to 4:1, plus free board. For example, depth 1.5 m → plan area = 2.40/1.5 = 1.6 m² → e.g., 1.8 m × 0.9 m (L:B = 2:1), with ~0.3 m free board above liquid level.

(a) Sludge volume after thickening

For sludge with constant solids mass and approximately constant specific gravity, volume varies inversely with the solids fraction:

$$\frac{V_2}{V_1} = \frac{P_1}{P_2}$$

where $P$ = percentage of solids (= 100 − moisture).

Initial solids: $P_1 = 100 - 96 = 4\%$. Final solids: $P_2 = 100 - 92 = 8\%$.

$$V_2 = V_1 \times \frac{P_1}{P_2} = 40 \times \frac{4}{8} = 20\ \text{m}^3$$

Thickened sludge volume = 20 m³ (halved, since solids concentration doubled).

Check: Dry solids in initial sludge = $40 \times 0.04 = 1.6\ \text{m}^3$ equivalent of solids; thickened = $20 \times 0.08 = 1.6$ — conserved.

(b) Anaerobic sludge digestion — objectives and products

Objectives:

• Stabilisation of organic solids — convert putrescible organic matter to stable end products, reducing odour and putrescibility.
• Volume and mass reduction — destruction of volatile solids reduces the quantity of sludge requiring disposal.
• Pathogen reduction — destroys many disease-causing organisms, improving safety for disposal/reuse.
• Improved dewaterability of the digested sludge and energy recovery through biogas.

Main end products: Biogas — primarily methane (CH₄, ~60–70%) and carbon dioxide (CO₂, ~30–40%), with traces of $H_2S$, $N_2$ — plus a stabilised digested sludge (digestate) and supernatant liquor.

(a) Levels of wastewater treatment

(b) Chlorination

Chlorination is the addition of chlorine (as $Cl_2$ gas, hypochlorite, or bleaching powder) to disinfect treated effluent by destroying pathogenic micro-organisms. Chlorine reacts with water:

$$Cl_2 + H_2O \rightarrow HOCl + HCl$$ $$HOCl \rightleftharpoons H^+ + OCl^-$$

Hypochlorous acid ($HOCl$) is the most effective germicidal form.

Chlorine demand: the amount of chlorine consumed in reacting with reducing agents, organic matter and ammonia in the wastewater before a free residual is available.

$$\text{Chlorine demand} = \text{Chlorine dose} - \text{Chlorine residual}$$

Breakpoint chlorination: As chlorine is added, it first reacts with reducing compounds, then with ammonia to form chloramines (combined residual), which rise to a peak and are then destroyed by further chlorine (oxidised to $N_2$). The breakpoint is the dose at which the combined residual drops to a minimum; beyond it, any added chlorine appears as free available residual chlorine, which gives rapid and reliable disinfection. Dosing past the breakpoint ensures a free chlorine residual for effective pathogen kill.

(a) Types of sewerage systems

(b) Why both minimum and maximum velocities are specified

• Minimum (self-cleansing) velocity (≈ 0.6–0.9 m/s): ensures that suspended solids and grit do not settle and deposit inside the sewer, which would cause silting, blockage and odour. The sewer must reach this velocity at least once a day at the design minimum flow.
• Maximum (non-scouring) velocity (≈ 2.5–3.0 m/s): limits erosion/abrasion of the sewer invert by grit and high-velocity flow, preventing damage to the pipe material and extending its life.

Thus the design velocity is kept within this self-cleansing-to-non-scouring band for reliable, durable operation.