BE Civil Engineering (IOE, TU) Water Supply Engineering (IOE, CE 651) Question Paper 2078 Nepal

(a) Population forecasting

First tabulate the per-decade increases.

Number of future decades from 2078 to 2098 = $n = 2$.

Arithmetic Increase Method

Mean increment:

$$\bar{x} = \frac{6300+8600+9700+12800}{4} = \frac{37400}{4} = 9350$$ $$P_{2098} = P_{2078} + n\,\bar{x} = 55600 + 2(9350) = 55600 + 18700 = 74{,}300$$

Arithmetic estimate ≈ 74,300.

Geometric Increase Method

Mean geometric growth ratio (geometric mean of the four ratios):

$$r = (1.3462 \times 1.3510 \times 1.2931 \times 1.2991)^{1/4}$$

Product = $1.3462\times1.3510 = 1.8187$; $1.8187\times1.2931 = 2.3520$; $2.3520\times1.2991 = 3.0556$.

$$r = (3.0556)^{1/4} = 1.3219$$ $$P_{2098} = P_{2078}\,r^{\,n} = 55600 \times (1.3219)^2 = 55600 \times 1.7474 = 97{,}154$$

Geometric estimate ≈ 97,150.

(b) Maximum daily demand (geometric population)

Design population $P = 97{,}150$, per-capita demand $q = 135$ lpcd.

Average daily demand:

$$Q_{avg} = \frac{97150 \times 135}{10^6} = \frac{13{,}115{,}250}{10^6} = 13.12\ \text{MLD}$$

Maximum daily demand (factor 1.8):

$$Q_{max} = 1.8 \times 13.12 = 23.61\ \text{MLD}$$

Maximum daily demand ≈ 23.6 MLD.

(a) Stokes' law settling velocity

Assumptions: particle is a smooth rigid sphere; settles discretely (no flocculation); laminar flow around particle (Reynolds number $R_e < 1$); fluid is quiescent; terminal (constant) velocity reached so net force = 0.

At terminal velocity, gravity (submerged weight) = drag force.

Submerged weight:

$$F_g = \frac{\pi}{6} d^3 (\rho_s - \rho)g$$

Stokes drag (laminar):

$$F_d = 3\pi \mu d\, v_s$$

Equating $F_g = F_d$:

$$3\pi \mu d\, v_s = \frac{\pi}{6} d^3 (\rho_s-\rho)g$$ $$\boxed{\,v_s = \frac{g\,(\rho_s-\rho)\,d^2}{18\,\mu}\,}$$

where $d$ = particle diameter, $\rho_s,\rho$ = particle and water densities, $\mu$ = dynamic viscosity.

(b) Tank design

Flow $Q = 18$ MLD $= 18\times10^3\ \text{m}^3/\text{day}$.

$$Q = \frac{18000}{86400} = 0.2083\ \text{m}^3/\text{s}$$

Design settling velocity $v_s = 0.30\ \text{mm/s} = 0.30\times10^{-3}\ \text{m/s} = 3.0\times10^{-4}\ \text{m/s}$.

For 100% removal of this particle, SOR (overflow rate) = $v_s$, so required surface area:

$$A = \frac{Q}{v_s} = \frac{0.2083}{3.0\times10^{-4}} = 694.4\ \text{m}^2$$

Surface overflow rate:

$$SOR = \frac{Q}{A} = v_s = 3.0\times10^{-4}\ \text{m/s} = 0.30\ \text{mm/s}$$

Expressed per day: $SOR = \dfrac{18000\ \text{m}^3/\text{day}}{694.4\ \text{m}^2} = 25.9\ \text{m}^3/\text{m}^2/\text{day}$ (typical: 20–40).

Dimensions with $L = 4B$:

$$A = L\times B = 4B^2 = 694.4 \Rightarrow B^2 = 173.6 \Rightarrow B = 13.18\ \text{m}$$ $$L = 4B = 52.7\ \text{m}$$

Adopt B = 13.2 m, L = 52.8 m, depth = 3.0 m.

Detention time check: Volume $V = L\times B\times H = 52.8\times13.2\times3.0 = 2090.9\ \text{m}^3$.

$$t = \frac{V}{Q} = \frac{2090.9}{0.2083} = 10{,}038\ \text{s} = 2.79\ \text{h}$$

Detention time ≈ 2.8 h (within the typical 2–4 h range). Design satisfactory.

(a) Filter area and number of units

Design flow to be filtered must include wash-water allowance and downtime.

Gross water to be produced/passed (accounting for 4% wash usage), and effective filtering time reduced by backwash downtime.

Backwash downtime per filter per day = 2 × 30 min = 60 min = 1 h. So effective operating time = 24 − 1 = 23 h/day.

Quantity of water to be filtered (including 4% extra for washing):

$$Q_{filt} = 24 \times 1.04 = 24.96\ \text{MLD} = 24.96\times10^6\ \text{L/day}$$

Volume filtered per m² in 23 effective hours:

$$= 5000\ \text{L/m}^2/\text{h} \times 23\ \text{h} = 115{,}000\ \text{L/m}^2/\text{day}$$

Required total area:

$$A = \frac{24.96\times10^6}{115000} = 217.0\ \text{m}^2$$

Area of one unit = $5\times6 = 30\ \text{m}^2$.

$$\text{No. of working units} = \frac{217.0}{30} = 7.23 \approx 8\ \text{units}$$

Provide 8 working units + 1 standby = 9 filter units of 5 m × 6 m each.

Provided working area = $8\times30 = 240\ \text{m}^2 > 217\ \text{m}^2$ — OK.

(b) Rapid vs slow sand filter (any three)

Three key differences: filtration rate, cleaning method (backwash vs scraping), and need for prior coagulation.

(a) Distribution layouts

Dead-end (tree) system: a main pipe runs through the centre, sub-mains branch off, and laterals branch from sub-mains; flow has a single path to any point (branches terminate as dead ends).

• Advantage: simple design, fewer valves, low initial cost; discharge/pressure easy to compute.
• Disadvantage: water stagnates at dead ends (poor quality); during repair a large area loses supply (no alternative path).

Grid-iron (interconnected) system: mains and sub-mains are interconnected so water can reach any point by more than one route, with no dead ends.

• Advantage: water circulates freely (no stagnation); during repair only a small portion is isolated; better for firefighting.
• Disadvantage: more pipe length and valves → higher cost; network analysis is more complex (requires Hardy-Cross/iterative methods).

(b) Hardy-Cross, first iteration

Loop A-B-C-D-A. Take clockwise = path through B positive, path through D negative.

Initial: $Q_1 = +70$ L/s (A→B→C, clockwise), $Q_2 = -50$ L/s (A→D→C is counter-clockwise around the loop).

Compute $h_f = KQ^2$ (signed) and $2|KQ|$:

Flow correction:

$$\Delta Q = -\frac{\sum KQ|Q|}{\sum 2K|Q|} = -\frac{8.80}{3.680} = -2.39\ \text{L/s}$$

Apply $\Delta Q$ to each pipe (add to the signed flow):

• Pipe 1: $Q_1' = 70 + (-2.39) = 67.61$ L/s (A→B→C)
• Pipe 2: $Q_2' = 50 - (-2.39) = 52.39$ L/s (A→D→C)

(Check continuity: $67.61 + 52.39 = 120$ L/s = inflow. ✓)

After first iteration: pipe 1 ≈ 67.6 L/s, pipe 2 ≈ 52.4 L/s. Since $|\Delta Q| = 2.39$ L/s is not yet negligible, iteration would continue.

(a) Total pumping head

Discharge $Q = 45$ L/s $= 0.045\ \text{m}^3/\text{s}$. Diameter $D = 0.25$ m. Pipe area:

$$A = \frac{\pi}{4}D^2 = \frac{\pi}{4}(0.25)^2 = 0.04909\ \text{m}^2$$

Velocity:

$$V = \frac{Q}{A} = \frac{0.045}{0.04909} = 0.9167\ \text{m/s}$$

Friction head (Darcy-Weisbach):

$$h_f = \frac{fLV^2}{2gD} = \frac{0.02\times300\times(0.9167)^2}{2\times9.81\times0.25}$$ $$= \frac{0.02\times300\times0.8403}{4.905} = \frac{5.0419}{4.905} = 1.028\ \text{m}$$

Minor losses = 15% of $h_f$ = $0.15\times1.028 = 0.154\ \text{m}$.

Total head:

$$H = H_{static} + h_f + h_{minor} = 28 + 1.028 + 0.154 = 29.18\ \text{m}$$

Total head ≈ 29.2 m.

(b) Power and energy

Water (output) power:

$$P_{water} = \rho g Q H = 1000\times9.81\times0.045\times29.18 = 12{,}882\ \text{W} = 12.88\ \text{kW}$$

Input power (overall efficiency 72%):

$$P_{input} = \frac{P_{water}}{\eta} = \frac{12.88}{0.72} = 17.89\ \text{kW}$$

Input power ≈ 17.9 kW.

Monthly energy (16 h/day × 30 days = 480 h):

$$E = 17.89\ \text{kW} \times 480\ \text{h} = 8{,}587\ \text{kWh}$$

Monthly energy consumption ≈ 8,590 kWh.

(a) Breakpoint chlorination

Breakpoint chlorination is the application of chlorine to water in a dose sufficient to satisfy all chlorine demand (reactions with ammonia, organic matter and other reducing agents), past the point where combined-residual chloramines are destroyed, so that any further chlorine added appears as free available chlorine residual.

Curve description (residual vs applied dose):

Residual
  |          free residual (slope 1)
  |                       /
  |      chloramine     /
  |       hump        /
  |       /\         /
  |      /  \       /
  |     /    \__   /  <- breakpoint (minimum)
  |    /        \ /
  |___/__________V________________ Applied dose
   demand   destruction of
   zone     chloramines

Region 1: chlorine consumed by reducing agents (no residual). Region 2: residual rises as chloramines form. Region 3: residual falls as chloramines are oxidised/destroyed, reaching the breakpoint (minimum). Region 4: beyond breakpoint, free residual increases linearly.

(b) Chlorine dose

Applied dose = chlorine demand + required free residual:

$$C = 2.4 + 0.5 = 2.9\ \text{mg/L}$$

Flow $Q = 12$ MLD $= 12\times10^6\ \text{L/day}$.

Mass of chlorine per day:

$$M = C \times Q = 2.9\ \frac{\text{mg}}{\text{L}} \times 12\times10^6\ \frac{\text{L}}{\text{day}} = 34.8\times10^6\ \text{mg/day}$$ $$= 34.8\ \text{kg/day}$$

Daily chlorine dose ≈ 34.8 kg/day.

(Shortcut: kg/day = dose in mg/L × flow in MLD = 2.9 × 12 = 34.8.)

(a) Coagulation-flocculation mechanism

Colloidal turbidity particles carry negative surface charges that keep them mutually repelled and non-settleable. Coagulation is the rapid addition and mixing of a coagulant (e.g. alum) that supplies trivalent cations ($\text{Al}^{3+}$) which neutralise the surface charge (destabilise the colloids) and form gelatinous hydroxide $\text{Al(OH)}_3$ flocs. Flocculation is the subsequent gentle stirring that allows the destabilised particles to collide and agglomerate into larger, heavier, settleable flocs (sweep-floc / enmeshment). Alum reacts with natural alkalinity to form the $\text{Al(OH)}_3$ floc.

(b) Alum and alkalinity

Alum requirement:

$$M = \text{dose} \times \text{flow} = 28\ \text{mg/L} \times 15\ \text{MLD} = 420\ \text{kg/day}$$

(Check: $28\times10^{-6}\,\text{kg/L} \times 15\times10^{6}\,\text{L/day} = 420$ kg/day.)

Alum requirement = 420 kg/day.

Alkalinity consumed:

$$= 0.45 \times 28 = 12.6\ \text{mg/L as CaCO}_3$$

Alkalinity consumed = 12.6 mg/L (as CaCO₃). If natural alkalinity is below this, lime/soda must be added.

(a) Sources of water

Classification:

1. Surface sources – rivers/streams, natural ponds & lakes, impounded reservoirs, springs (surface-fed).
2. Sub-surface (ground) sources – infiltration galleries, springs (artesian), wells (open/dug wells, shallow & deep tube wells), aquifers.

Sometimes a third class — rainwater (rooftop harvesting) — is added, important for rural/hill areas of Nepal.

Merit of surface over groundwater: surface sources usually provide large quantities sufficient for big towns and are easy to develop/abstract; (groundwater is limited in yield though generally better in quality and needs less treatment).

(b) Water-quality parameters

Permissible limits (Nepal NDWQS / WHO guidance):

• Turbidity: 5 NTU (desirable ≤ 1 NTU for effective disinfection).
• Total coliform: 0 (nil) per 100 mL in any drinking-water sample (must be absent).

Balancing storage capacity

Total daily supply pumped = daily demand = 6.0 MLD = 6,000 m³/day.

Pumping is uniform over 12 h, so pumping rate:

$$= \frac{6000}{12} = 500\ \text{m}^3/\text{h (only during 06:00–18:00)}$$

Consumption (demand) is uniform over 24 h:

$$= \frac{6000}{24} = 250\ \text{m}^3/\text{h (all 24 h)}$$

During pumping hours (12 h, 06:00–18:00): inflow 500, outflow 250 → net surplus = $+250\ \text{m}^3/\text{h}$. Accumulated surplus over 12 h $= 250\times12 = 3{,}000\ \text{m}^3$.

During non-pumping hours (12 h, 18:00–06:00): inflow 0, outflow 250 → net deficit $= -250\ \text{m}^3/\text{h}$. Over 12 h $= 3{,}000\ \text{m}^3$ drawn down.

The required balancing storage = maximum cumulative surplus (= maximum cumulative deficit):

$$V_{bal} = 3{,}000\ \text{m}^3$$

As a fraction of daily supply:

$$\frac{3000}{6000} = 0.50 = 50\%\ \text{of the day's demand.}$$

Balancing storage ≈ 3,000 m³ (50% of daily demand). A breakdown/fire reserve would be added to this for the total reservoir capacity.

(a) Gravity-flow rural water supply scheme

A gravity-flow scheme conveys water from a higher-elevation source (typically a hill spring or stream) to a lower-elevation settlement entirely by gravity, without pumping, using the natural head difference. It is the most common system for the hills of Nepal because it has no energy cost and is simple to operate.

Main components (source → tap):

1. Intake / source – spring source protection or stream intake (sometimes a small weir).
2. Collection / sedimentation chamber – removes coarse sediment.
3. Transmission (conveyance) main – HDPE/GI pipeline from source to tank.
4. Break-pressure tanks (BPT) – placed along steep stretches to limit static head and protect pipes.
5. Reservoir / service (storage) tank (RVT) – balances supply and demand.
6. Distribution pipeline – carries water to clusters.
7. Tapstands / public standposts (or household connections) with valves.

(b) Household point-of-use treatment (any two)

1. Boiling – kills pathogens effectively; limitation: high fuel/energy use and does not remove turbidity or chemicals.
2. SODIS (solar disinfection in PET bottles) – cheap, no chemicals; limitation: needs sunny weather and clear (low-turbidity) water, small batch volumes.
3. Chlorination (Piyush/sodium hypochlorite drops) – simple residual protection; limitation: taste/odour issues and ineffective if water is turbid.
4. Ceramic / candle / biosand filters – remove turbidity and bacteria; limitation: slow flow, need regular cleaning, limited virus removal.

Fire demand (Kuichling formula)

Population in thousands: $P = 80{,}000/1000 = 80$.

$$Q = 3182\sqrt{P} = 3182\times\sqrt{80} = 3182\times8.944 = 28{,}460\ \text{L/min}$$

Convert to L/day (continuous basis for capacity check):

$$Q = 28{,}460\ \text{L/min} \times 60\times24\ \text{min/day} = 28{,}460 \times 1440$$ $$= 40{,}982{,}400\ \text{L/day} = 40.98\ \text{MLD}$$

Fire demand ≈ 28,460 L/min ≈ 41.0 MLD (if sustained continuously).

Comment / how it is provided: Fire demand is a short-duration, high-intensity draft, not a continuous load. It is therefore not added to the average daily demand directly; instead a fire reserve volume is kept in the service reservoir and the distribution mains are sized to deliver the required fire flow at adequate residual pressure during the (typically a few hours) fire event. In design the governing demand is usually the greater of: (i) maximum daily demand, or (ii) average daily demand + fire demand (coincident-draft check). For an 80,000-population city the sustained 41 MLD figure simply indicates the instantaneous capacity the mains/hydrants must momentarily supply, met from stored reserve rather than continuous production.