BE Civil Engineering (IOE, TU) Sanitary Engineering (IOE, CE 654) Question Paper 2077 Nepal

Given: Population $p = 18{,}000$, water supply $= 135$ L/c/day, 80% reaches sewer, infiltration $= 18{,}000$ L/ha/day over $45$ ha, $n = 0.013$, slope $S = 1/600$.

(a) Average DWF

Sanitary wastewater:

$$Q_{san} = 18{,}000 \times 135 \times 0.80 = 1{,}944{,}000 \text{ L/day}$$

Infiltration:

$$Q_{inf} = 18{,}000 \times 45 = 810{,}000 \text{ L/day}$$

Total average flow:

$$Q_{avg} = 1{,}944{,}000 + 810{,}000 = 2{,}754{,}000 \text{ L/day}$$

Convert to L/s ($86{,}400$ s/day):

$$Q_{avg} = \frac{2{,}754{,}000}{86{,}400} = \mathbf{31.88 \text{ L/s}}$$

(b) Peak sewage flow

Apply the peak factor only to the sanitary component (infiltration is taken as steady). With $p = 18$ (thousands):

$$P = \frac{5}{18^{0.2}} = \frac{5}{1.7826} = 2.805$$

Peak sanitary flow:

$$Q_{san} = \frac{1{,}944{,}000}{86{,}400} = 22.50 \text{ L/s}$$ $$Q_{peak,san} = 2.805 \times 22.50 = 63.11 \text{ L/s}$$

Add steady infiltration ($810{,}000/86{,}400 = 9.375$ L/s):

$$Q_{peak} = 63.11 + 9.375 = \mathbf{72.49 \text{ L/s}} \approx 0.0725 \text{ m}^3/\text{s}$$

(c) Sewer diameter (flowing full)

For a circular pipe flowing full, Manning's equation gives:

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

Combining:

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

Numerically, $\dfrac{\pi}{4\cdot 4^{2/3}} = \dfrac{3.1416}{4\times 2.5198} = 0.3117$ and $S^{1/2} = \sqrt{1/600} = 0.04082$:

$$0.0725 = 0.3117 \times \frac{0.04082}{0.013}\, D^{8/3} = 0.3117 \times 3.140\, D^{8/3} = 0.9788\, D^{8/3}$$ $$D^{8/3} = \frac{0.0725}{0.9788} = 0.07407$$ $$D = 0.07407^{3/8} = \mathbf{0.367 \text{ m}}$$

Provide the next commercial size $D = 0.40$ m (400 mm).

Velocity check at full flow for $D = 0.40$ m:

$$A = \frac{\pi (0.40)^2}{4} = 0.1257 \text{ m}^2, \quad R = \frac{0.40}{4} = 0.10 \text{ m}$$ $$V = \frac{1}{0.013}(0.10)^{2/3}(0.04082) = \frac{1}{0.013}\times 0.2154 \times 0.04082 = 0.676 \text{ m/s}$$

Since $V = 0.68$ m/s $> 0.6$ m/s, the self-cleansing velocity is satisfied. (Even the computed exact diameter 0.367 m would give a slightly higher velocity, so a 400 mm pipe at the design slope is acceptable.)

(a) Definitions and significance

• BOD (Biochemical Oxygen Demand): the mass of dissolved oxygen consumed per unit volume of sample by microorganisms while oxidizing the biodegradable organic matter under aerobic conditions, conventionally measured over 5 days at 20 °C.
• COD (Chemical Oxygen Demand): the oxygen equivalent of organic (and some inorganic) matter that is oxidized by a strong chemical oxidant (e.g. dichromate). COD ≥ BOD always, because chemical oxidation attacks both biodegradable and non-biodegradable matter.
• BOD/COD ratio: an index of biodegradability. A ratio $>0.5$ (typical of domestic sewage) indicates readily biodegradable waste amenable to biological treatment; a ratio $<0.3$ indicates a large refractory (industrial) fraction, requiring physico-chemical treatment.

(b) First-stage BOD kinetics

The first-order model: $BOD_t = L_0\,(1 - e^{-k_1 t})$.

(i) Ultimate BOD from $BOD_5$:

$$L_0 = \frac{BOD_5}{1 - e^{-k_1 \times 5}} = \frac{220}{1 - e^{-0.23\times 5}} = \frac{220}{1 - e^{-1.15}}$$ $$e^{-1.15} = 0.3166 \Rightarrow L_0 = \frac{220}{0.6834} = \mathbf{321.9 \text{ mg/L}}$$

(ii) BOD exerted after 8 days and BOD remaining:

$$BOD_8 = 321.9\,(1 - e^{-0.23\times 8}) = 321.9\,(1 - e^{-1.84})$$ $$e^{-1.84} = 0.1588 \Rightarrow BOD_8 = 321.9 \times 0.8412 = 270.8 \text{ mg/L (exerted)}$$

BOD remaining (unoxidized) after 8 days:

$$L_8 = L_0\,e^{-k_1 t} = 321.9 \times 0.1588 = \mathbf{51.1 \text{ mg/L}}$$

(c) Temperature correction to 27 °C

$$k_{27} = k_{20}\,\theta^{(27-20)} = 0.23 \times 1.047^{7}$$ $$1.047^{7} = 1.3759 \Rightarrow k_{27} = 0.23 \times 1.3759 = \mathbf{0.3165 \text{ day}^{-1}}$$

$BOD_5$ at 27 °C (with $L_0 = 321.9$ mg/L unchanged):

$$BOD_{5,27} = 321.9\,(1 - e^{-0.3165\times 5}) = 321.9\,(1 - e^{-1.5825})$$ $$e^{-1.5825} = 0.2055 \Rightarrow BOD_{5,27} = 321.9 \times 0.7945 = \mathbf{255.8 \text{ mg/L}}$$

The higher temperature accelerates oxidation, so the same sample exerts a larger 5-day BOD.

Given: $Q = 6{,}000$ m³/day, SOR $= 30$ m³/m²/day, $t_d = 2.0$ h, $L:W = 4:1$.

(a) Surface area and dimensions

Surface area:

$$A = \frac{Q}{\text{SOR}} = \frac{6{,}000}{30} = 200 \text{ m}^2$$

With $L = 4W$: $A = L\times W = 4W^2 = 200 \Rightarrow W^2 = 50$.

$$W = \sqrt{50} = 7.07 \text{ m} \approx \mathbf{7.0 \text{ m}}, \quad L = 4\times 7.07 = 28.3 \text{ m} \approx \mathbf{28.5 \text{ m}}$$

Volume from detention time:

$$V = Q\times t_d = 6{,}000 \times \frac{2.0}{24} = 500 \text{ m}^3$$

Depth:

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

(liquid depth; add ~0.3 m freeboard).

Using the adopted dimensions ($28.5 \times 7.0 = 199.5$ m²) the design is consistent.

(b) Weir loading

Weir length $= W = 7.0$ m.

$$\text{Weir loading} = \frac{Q}{\text{weir length}} = \frac{6{,}000}{7.0} = \mathbf{857 \text{ m}^3/\text{m/day}}$$

This greatly exceeds $250$ m³/m/day, so a single end weir is unacceptable. A finger/launder weir arrangement is needed to give a length of at least $6{,}000/250 = 24$ m of weir crest to control short-circuiting and floc carry-over.

(c) Functions and additional design checks

Functions of primary sedimentation:

1. Remove $50$–$70\%$ of suspended solids and $25$–$40\%$ of BOD by gravity settling, reducing the load on downstream biological units.
2. Remove floating matter (scum/grease) and provide a point for raw-sludge collection.

Other governing design checks:

1. Detention time (1.5–2.5 h) to allow discrete and flocculent settling.
2. Horizontal (displacement) velocity / weir loading to avoid scour and floc carry-over; also Reynolds and Froude numbers for hydraulic stability.

Given: $Q = 5{,}000$ m³/day, $S_0 = 200$ mg/L, $S = 20$ mg/L, $X = 3{,}000$ mg/L, $\theta_c = 8$ d, $Y = 0.6$, $k_d = 0.06$ d⁻¹.

(a) Aeration tank volume and HRT

The completely-mixed design equation:

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

Keep concentrations in mg/L (= g/m³) and flow in m³/day so units of mass cancel:

$$V = \frac{8 \times 5{,}000 \times 0.6 \times (200 - 20)}{3{,}000 \times (1 + 0.06\times 8)}$$

Numerator $= 8 \times 5{,}000 \times 0.6 \times 180 = 4{,}320{,}000$. Denominator $= 3{,}000 \times (1 + 0.48) = 3{,}000 \times 1.48 = 4{,}440$.

$$V = \frac{4{,}320{,}000}{4{,}440} = \mathbf{973.0 \text{ m}^3}$$

Hydraulic retention time:

$$\tau = \frac{V}{Q} = \frac{973.0}{5{,}000} = 0.1946 \text{ day} = \mathbf{4.67 \text{ hours}}$$

(b) Waste activated sludge production

Observed yield:

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

Mass of sludge produced per day:

$$P_x = Y_{obs}\,Q\,(S_0 - S) = 0.4054 \times 5{,}000 \times (180\ \text{g/m}^3)$$ $$P_x = 0.4054 \times 5{,}000 \times 0.180\ \text{kg/m}^3 = \mathbf{364.9 \text{ kg VSS/day}}$$

(Check via $P_x = XV/\theta_c = 3000\times973.0/8 \times 10^{-3} = 364.9$ kg/day — consistent.)

(c) F/M ratio and volumetric loading

F/M ratio:

$$\frac{F}{M} = \frac{Q\,S_0}{X\,V} = \frac{5{,}000 \times 200}{3{,}000 \times 973.0} = \frac{1{,}000{,}000}{2{,}919{,}000} = \mathbf{0.343 \text{ kg BOD/kg MLSS/day}}$$

Volumetric BOD loading:

$$L_v = \frac{Q\,S_0}{V} = \frac{5{,}000 \times 0.200\ \text{kg/m}^3}{973.0} = \frac{1{,}000\ \text{kg/day}}{973.0\ \text{m}^3} = \mathbf{1.03 \text{ kg BOD/m}^3/\text{day}}$$

Both values lie in the conventional activated-sludge range (F/M $0.2$–$0.5$, $L_v$ $0.3$–$1.0$ kg/m³/day), confirming a sound design.

Given: rate $= 0.45$ kg/c/day, population $= 120{,}000$, loose density $= 250$ kg/m³, vehicle volume $= 12$ m³, in-vehicle compacted density $= 400$ kg/m³, landfill density $= 700$ kg/m³.

(a) Daily mass and volume

Daily mass:

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

Volume at loose (as-collected) density:

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

(b) Number of vehicle trips

Mass carried per trip at $400$ kg/m³ compaction:

$$m_{trip} = 12 \times 400 = 4{,}800 \text{ kg/trip}$$

Trips required:

$$N = \frac{54{,}000}{4{,}800} = 11.25 \Rightarrow \mathbf{12 \text{ trips/day}}$$

(round up to clear all waste).

(c) Annual landfill airspace and soil-cover functions

Daily landfilled volume at $700$ kg/m³:

$$V_{LF,day} = \frac{54{,}000}{700} = 77.14 \text{ m}^3/\text{day}$$

Annual airspace consumed by waste:

$$V_{LF,year} = 77.14 \times 365 = \mathbf{28{,}157 \text{ m}^3/\text{year}}$$

If daily soil cover is taken as roughly 20% extra of cell volume, total airspace ≈ $28{,}157 \times 1.2 \approx 33{,}789$ m³/year.

Functions of daily cover:

1. Controls disease vectors (flies, rodents), wind-blown litter and odour, and reduces fire risk.
2. Limits rainwater infiltration into the cell (reducing leachate) and improves the working surface for vehicles.

1. Manhole

• Purpose: provides access for inspection, cleaning and jointing; located at changes of direction, gradient, size and at junctions, and at regular intervals (≈ 30 m for small, up to 90 m for large sewers) on straight runs.
• Design considerations: internal size large enough for a worker (≈ 0.9–1.2 m chamber), benching sloped to the channel, step-irons or ladder, watertight cover to carry traffic load, and a flow channel (invert) matching the sewer.

        ____cover____
       |   frame    |
  GL --|------------|-- GL
       |  shaft     |
       | (corbel)   |
       |   working  |
       |   chamber  |
  =====|  benching  |=====
   in -->  channel  --> out

2. Drop manhole

• Purpose: used where an incoming sewer is much higher than the outgoing one (steep ground) to lower the flow without high velocity inside the chamber or splashing on workers.
• Design considerations: a vertical drop pipe (down-pipe) connects the high inlet to the channel; an access/cleaning eye is provided at the top of the drop; used when the drop exceeds about 0.6 m.

3. Inverted siphon (depressed sewer)

• Purpose: carries the sewer beneath an obstruction (stream, road cutting, utility) by dipping below the hydraulic grade line; flows full under pressure.
• Design considerations: multiple parallel barrels sized so that a self-cleansing velocity (≥ 0.9 m/s) is maintained at minimum flow in at least one barrel; inlet/outlet chambers with penstocks for isolation and cleaning; provision against sediment deposition.

Given: $D = 0.30$ m, $S = 1/400 = 0.0025$, $n = 0.013$.

(a) Full-flow conditions

$$A = \frac{\pi D^2}{4} = \frac{\pi (0.30)^2}{4} = 0.07069 \text{ m}^2, \quad R = \frac{D}{4} = 0.075 \text{ m}$$

Full-flow velocity:

$$V = \frac{1}{n} R^{2/3} S^{1/2} = \frac{1}{0.013}(0.075)^{2/3}(0.0025)^{1/2}$$ $$(0.075)^{2/3} = 0.1779, \quad (0.0025)^{1/2} = 0.05$$ $$V = \frac{1}{0.013}\times 0.1779 \times 0.05 = \frac{0.008895}{0.013} = \mathbf{0.684 \text{ m/s}}$$

Full-flow discharge:

$$Q = A\,V = 0.07069 \times 0.684 = 0.04835 \text{ m}^3/\text{s} = \mathbf{48.35 \text{ L/s}}$$

(b) Partial flow at $q = 0.40\,Q$

Discharge:

$$q = 0.40 \times 48.35 = \mathbf{19.34 \text{ L/s}}$$

From the hydraulic-elements chart, $q/Q = 0.40 \Rightarrow d/D \approx 0.45$, so the flow depth is:

$$d = 0.45 \times 300 = \mathbf{135 \text{ mm}}$$

The corresponding velocity ratio is $v/V \approx 0.90$:

$$v = 0.90 \times 0.684 = \mathbf{0.616 \text{ m/s}}$$

Since $v = 0.62$ m/s $\geq 0.6$ m/s, the sewer remains self-cleansing even at $40\%$ of full flow.

Given: $N = 8$ persons, flow $= 90$ L/c/day, $t_d = 24$ h $= 1$ day, sludge storage $= 40$ L/person/year, desludging interval $= 3$ years, liquid depth $= 1.5$ m, $L:W = 3:1$.

Step 1 — Liquid (sewage) volume

$$V_{liquid} = N \times q \times t_d = 8 \times 90 \times 1 = 720 \text{ L} = 0.72 \text{ m}^3$$

Step 2 — Sludge and scum storage volume

$$V_{sludge} = N \times 40 \times 3 = 8 \times 40 \times 3 = 960 \text{ L} = 0.96 \text{ m}^3$$

Step 3 — Total effective capacity

$$V = V_{liquid} + V_{sludge} = 0.72 + 0.96 = \mathbf{1.68 \text{ m}^3}$$

Step 4 — Plan area and dimensions (liquid depth 1.5 m)

$$A = \frac{V}{d} = \frac{1.68}{1.5} = 1.12 \text{ m}^2$$

With $L = 3W$: $A = 3W^2 = 1.12 \Rightarrow W^2 = 0.3733$.

$$W = 0.611 \text{ m} \Rightarrow \text{adopt } \mathbf{W = 0.75 \text{ m}}, \quad L = 3W = \mathbf{2.25 \text{ m}}$$

(A minimum practical width of 0.75 m is adopted for desludging access; this gives $A = 1.69$ m² and effective volume $= 1.69 \times 1.5 = 2.53$ m³, comfortably exceeding the required 1.68 m³.)

Step 5 — Provide freeboard

Total depth $= 1.5$ m liquid $+ 0.3$ m freeboard $= \mathbf{1.8 \text{ m}}$.

Adopted septic tank: $2.25\ \text{m (L)} \times 0.75\ \text{m (W)} \times 1.8\ \text{m (total depth)}$, with effluent led to a soak pit / leach field.

(a) Objectives and sequence

Objectives of sludge treatment:

• Reduce water content (volume) to cut handling and disposal cost.
• Stabilize the organics to reduce putrescibility and odour.
• Reduce pathogens for safe reuse/disposal.
• Recover useful by-products (biogas, nutrients) where feasible.

Usual sequence of unit operations:

$$\text{Thickening} \rightarrow \text{Stabilization (digestion)} \rightarrow \text{Conditioning} \rightarrow \text{Dewatering} \rightarrow \text{Final disposal/Reuse}$$

(b) Thickening mass balance

Solids are conserved. Using $\rho \approx 1000$ kg/m³ for the slurry and concentration by mass:

Mass of dry solids in raw sludge:

$$M_s = 30 \text{ m}^3 \times 1000 \text{ kg/m}^3 \times 0.04 = 1{,}200 \text{ kg/day}$$

Thickened-sludge volume at $7\%$ solids (same solids mass):

$$V_2 = \frac{M_s}{\rho \times 0.07} = \frac{1{,}200}{1000 \times 0.07} = \frac{1{,}200}{70} = \mathbf{17.14 \text{ m}^3/\text{day}}$$

A quicker route is the inverse-ratio rule $V_2 = V_1\,(P_1/P_2)$:

$$V_2 = 30 \times \frac{4}{7} = 17.14 \text{ m}^3/\text{day} \ ✓$$

Percentage volume reduction:

$$\frac{30 - 17.14}{30} \times 100 = \frac{12.86}{30}\times 100 = \mathbf{42.9\%}$$

Thickening from 4% to 7% therefore cuts the sludge volume by about 43%.

(a) Tertiary (advanced) treatment

Tertiary treatment is the polishing stage applied after secondary (biological) treatment to remove residual contaminants that secondary treatment cannot adequately remove, in order to meet stringent discharge or reuse standards.

Three common processes and their targets:

1. Filtration (sand / dual-media / membrane): removes residual suspended solids and turbidity.
2. Nutrient removal — nitrification–denitrification / biological or chemical P removal: removes nitrogen and phosphorus to control eutrophication.
3. Disinfection (chlorination, UV, ozonation): destroys pathogenic microorganisms.

(Activated-carbon adsorption for refractory organics/colour is another valid example.)

(b) Chlorine demand and consumption

Applied chlorine dose $= 8$ mg/L over $Q = 4{,}000$ m³/day.

Daily mass of chlorine applied:

$$M_{applied} = 8 \text{ mg/L} \times 4{,}000 \text{ m}^3/\text{day}$$

Using $1\ \text{mg/L} \times 1\ \text{m}^3 = 1\ \text{g}$:

$$M_{applied} = 8 \times 4{,}000 = 32{,}000 \text{ g/day} = \mathbf{32.0 \text{ kg/day}}$$

Chlorine demand = applied dose − residual = $8 - 0.5 = 7.5$ mg/L.

Daily mass of chlorine consumed (demand exerted):

$$M_{consumed} = 7.5 \times 4{,}000 = 30{,}000 \text{ g/day} = \mathbf{30.0 \text{ kg/day}}$$

The remaining $0.5$ mg/L ($= 2.0$ kg/day) leaves as free residual that maintains disinfecting capacity in the contact tank / receiving line.

Given: $W = 90$ kg/day, $V = 250$ m³, $R = 1.5$.

(a) NRC efficiency

Recirculation factor:

$$F = \frac{1 + R}{(1 + 0.1R)^2} = \frac{1 + 1.5}{(1 + 0.15)^2} = \frac{2.5}{1.3225} = 1.8904$$

Loading term inside the root:

$$\frac{W}{V\,F} = \frac{90}{250 \times 1.8904} = \frac{90}{472.6} = 0.19044$$ $$\sqrt{0.19044} = 0.4364$$

Efficiency:

$$E = \frac{100}{1 + 0.4432 \times 0.4364} = \frac{100}{1 + 0.19341} = \frac{100}{1.19341} = \mathbf{83.8\%}$$

The single-stage trickling filter removes about 84% of the applied BOD.

(b) Trickling filter vs activated sludge

Advantages of trickling filters:

1. Lower energy use and simpler operation — no fine control of MLSS, sludge return or dissolved oxygen.
2. More robust to shock/toxic loads and intermittent operation; less skilled supervision needed.

Disadvantages of trickling filters:

1. Generally lower and less consistent effluent quality than well-run activated sludge; limited ability to meet very low BOD.
2. Problems with filter flies, odour, and clogging (ponding); larger land area and higher capital cost for media/underdrains.