BE Civil Engineering (IOE, TU) Water Supply Engineering (IOE, CE 651) Question Paper 2076 Nepal

(a) Population forecasting

Decadal data and increments

Number of future decades from 2010 to 2040 = $(2040-2010)/10 = 3$ decades.

Geometric increase method

Geometric mean of the growth ratios:

$$r = \sqrt[3]{1.2321 \times 1.2696 \times 1.2831}$$

Product $= 1.2321 \times 1.2696 \times 1.2831 = 2.00723$.

$$r = (2.00723)^{1/3} = 1.2615$$

So the average decadal growth rate $\approx 26.15\%$.

Projected population:

$$P_{2040} = P_{2010}\, r^{3} = 56{,}200 \times (1.2615)^{3}$$ $$(1.2615)^{3} = 2.0072$$ $$P_{2040} = 56{,}200 \times 2.0072 = 112{,}805 \approx \mathbf{112{,}800}$$

Incremental increase method

Average increase $\bar{x} = (6{,}500 + 9{,}300 + 12{,}400)/3 = 28{,}200/3 = 9{,}400$.

Average incremental increase $\bar{y} = (2{,}800 + 3{,}100)/2 = 5{,}900/2 = 2{,}950$.

For $n$ decades ahead:

$$P = P_{2010} + n\bar{x} + \frac{n(n+1)}{2}\bar{y}$$

With $n = 3$:

$$P_{2040} = 56{,}200 + 3(9{,}400) + \frac{3 \times 4}{2}(2{,}950)$$ $$= 56{,}200 + 28{,}200 + 6 \times 2{,}950$$ $$= 56{,}200 + 28{,}200 + 17{,}700 = \mathbf{102{,}100}$$

(b) Demand computation

Design population (geometric) $= 112{,}800$.

Average daily demand:

$$Q_{avg} = 112{,}800 \times 135\ \text{lpcd} = 15{,}228{,}000\ \text{L/day} = \mathbf{15.23\ MLD}$$

Maximum daily demand (factor 1.8):

$$Q_{max\,day} = 1.8 \times 15.23 = \mathbf{27.41\ MLD}$$

Comment

For a fast-growing town with increasing decadal growth ratios, the geometric increase method is generally more appropriate because growth is compounding (proportional to the existing base). The arithmetic/incremental methods tend to under-predict for rapidly expanding settlements, as confirmed here (102,100 vs 112,800).

(a) Surface area and detention time

Flow rate:

$$Q = 18\ \text{MLD} = 18 \times 10^{6}\ \text{L/day} = 18{,}000\ \text{m}^3/\text{day}$$ $$Q = \frac{18{,}000}{86{,}400} = 0.2083\ \text{m}^3/\text{s}$$

Surface overflow rate must equal the settling velocity:

$$v_s = 0.30\ \text{mm/s} = 0.30 \times 10^{-3}\ \text{m/s} = 3.0 \times 10^{-4}\ \text{m/s}$$

Required surface area:

$$A = \frac{Q}{v_s} = \frac{0.2083}{3.0 \times 10^{-4}} = 694.4\ \text{m}^2 \approx \mathbf{694\ m^2}$$

Tank volume:

$$V = A \times H = 694.4 \times 3.5 = 2{,}430.6\ \text{m}^3$$

Theoretical detention time:

$$t = \frac{V}{Q} = \frac{2{,}430.6}{0.2083} = 11{,}669\ \text{s} = \mathbf{3.24\ hours}$$

(b) Dimensions and horizontal velocity

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

$$4B^2 = 694.4 \Rightarrow B^2 = 173.6 \Rightarrow B = 13.18\ \text{m} \approx \mathbf{13.2\ m}$$ $$L = 4 \times 13.18 = 52.7\ \text{m} \approx \mathbf{52.7\ m}$$

Cross-sectional flow area $= B \times H = 13.18 \times 3.5 = 46.13\ \text{m}^2$.

Horizontal velocity:

$$v_H = \frac{Q}{B \times H} = \frac{0.2083}{46.13} = 4.515 \times 10^{-3}\ \text{m/s} = \mathbf{4.52\ mm/s}$$

Since $4.52\ \text{mm/s} < 9\ \text{mm/s}$, the horizontal velocity is below the scouring limit, so re-suspension of settled solids is unlikely — the design is acceptable.

(c) Effect of short-circuiting factor

Real tanks never behave as ideal plug flow. Density currents, wind, thermal stratification and inlet/outlet turbulence cause some water to traverse the tank faster than the mean. The flow-through (short-circuiting) factor $\eta = t_{actual}/t_{theoretical}$ is typically 0.3–0.7. A lower $\eta$ means a portion of the flow has a shorter effective detention time than designed, reducing removal efficiency. Designers therefore provide baffles, good inlet diffusion, and adopt a higher detention time (or lower overflow rate) than the theoretical minimum to compensate.

(a) Filter area and number of units

Flow:

$$Q = 24\ \text{MLD} = 24{,}000\ \text{m}^3/\text{day}$$

Filtration rate:

$$5000\ \text{L/m}^2/\text{hr} = 5\ \text{m}^3/\text{m}^2/\text{hr} \times 24 = 120\ \text{m}^3/\text{m}^2/\text{day}$$

Total area:

$$A = \frac{24{,}000}{120} = 200\ \text{m}^2$$

Number of working units at 30 m² each:

$$n = \frac{200}{30} = 6.67 \Rightarrow 7\ \text{units (working)}$$

Adopt area per unit $= 200/7 = 28.6\ \text{m}^2$ ($\approx$ within the 30 m² limit). Providing one stand-by/extra unit for backwashing:

$$\text{Total} = 7 + 1 = \mathbf{8\ filter\ units}$$

(b) Wash-water requirement

Area per unit $\approx 28.6\ \text{m}^2$.

Wash water per unit per backwash:

$$= 28.6\ \text{m}^2 \times 0.6\ \text{m}^3/\text{m}^2/\text{min} \times 10\ \text{min} = 171.6\ \text{m}^3$$

With 7 working units backwashed once per day:

$$\text{Total wash water} = 7 \times 171.6 = 1{,}201.2\ \text{m}^3/\text{day}$$

Percentage of filtered water used:

$$\frac{1{,}201.2}{24{,}000} \times 100 = \mathbf{5.0\%}$$

This is within the typical acceptable range of 2–6 % for rapid sand filters.

(c) Operations in one filtration cycle

1. Filtration (service run) — raw/coagulated-settled water passes downward through the sand bed; runs until head loss reaches the limit (~2.5 m) or turbidity breaks through.
2. Backwashing — filtration stopped; water (sometimes with air scour) forced upward to fluidize and expand the bed, dislodging trapped floc.
3. Washing/rinsing — dirty wash water carried to the wash-water troughs and drained.
4. Filtering to waste (ripening) — initial filtrate after backwash run to waste until effluent turbidity is acceptable.
5. Filter returned to service; cycle repeats.

(a) Total dynamic head

Flow $Q = 40\ \text{L/s} = 0.040\ \text{m}^3/\text{s}$.

Pipe diameter $D = 0.25\ \text{m}$; area:

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

Velocity:

$$V = \frac{Q}{A} = \frac{0.040}{0.04909} = 0.8148\ \text{m/s}$$

Friction (Darcy–Weisbach) head loss:

$$h_f = \frac{fLV^2}{2gD} = \frac{0.020 \times 480 \times (0.8148)^2}{2 \times 9.81 \times 0.25}$$

Numerator $= 0.020 \times 480 \times 0.6639 = 6.374$. Denominator $= 2 \times 9.81 \times 0.25 = 4.905$.

$$h_f = \frac{6.374}{4.905} = 1.299\ \text{m}$$

Minor losses $= 0.15 \times 1.299 = 0.195\ \text{m}$.

Total dynamic head:

$$H = H_{static} + h_f + h_{minor} = 32 + 1.299 + 0.195 = \mathbf{33.49\ m}$$

(b) Power input

Water (useful) power:

$$P_{water} = \rho g Q H = 1000 \times 9.81 \times 0.040 \times 33.49 = 13{,}141\ \text{W} = 13.14\ \text{kW}$$

Power input at 70 % efficiency:

$$P_{input} = \frac{P_{water}}{\eta} = \frac{13.14}{0.70} = \mathbf{18.78\ kW}$$

(c) Monthly energy cost

Daily energy:

$$E = 18.78\ \text{kW} \times 16\ \text{hr} = 300.5\ \text{kWh/day}$$

Monthly (30 days):

$$E_{month} = 300.5 \times 30 = 9{,}014\ \text{kWh}$$

Cost:

$$= 9{,}014 \times 11 = \textbf{NRs 99{,}154 per month}$$

(a) Dead-end vs grid-iron systems

Dead-end (tree) system

• Advantages: simple design and easy to compute flows; cheaper (less pipe), suited to irregularly developed/old towns.
• Disadvantage: water at the far ends becomes stagnant (quality deteriorates); during repair a large area loses supply; limited fire-flow reliability.

Grid-iron (looped) system

• Advantages: water can reach any point from more than one direction, so no stagnation and good for firefighting; a single pipe break isolates only a small zone.
• Disadvantage: more pipe length and valves → higher cost; flow analysis is more complex (requires iterative methods).

(b) Hardy-Cross method

Principle: Successive approximation. Flows are first assumed in every pipe so that continuity at each node is satisfied (inflow = outflow). The head loss around each loop is then checked; a correction $\Delta Q$ is applied to all pipes of the loop and iterated until the loops balance.

Two governing conditions of a balanced network:

1. Continuity (junction law): at every node, $\sum Q_{in} = \sum Q_{out}$.
2. Energy (loop law): around every closed loop, the algebraic sum of head losses is zero, $\sum h_L = 0$.

With head loss expressed as $h_L = kQ^n$, the loop correction is:

$$\Delta Q = -\frac{\sum kQ^{n}}{n\sum k|Q|^{\,n-1}}$$

where $k$ = pipe resistance coefficient, $Q$ = assumed discharge (signed, clockwise positive), $n$ = exponent (1.85 for Hazen–Williams, 2 for Darcy), and the denominator uses absolute values.

(c) First correction

$$\Delta Q = -\frac{\sum kQ^{n}}{n\sum k Q^{\,n-1}} = -\frac{+6.0}{40}$$

(Note: the denominator $n\sum kQ^{n-1} = 40$ is already supplied including the factor $n$.)

$$\Delta Q = -0.15\ \text{(consistent units)}$$

So a correction of $\Delta Q = -0.15$ is applied to every pipe in the loop (subtracted from clockwise flows, added to anticlockwise), and the procedure is repeated until $\sum kQ^n \approx 0$.

(a) Daily alum requirement

$$\text{Mass} = \text{dose} \times Q = 28\ \text{mg/L} \times 20 \times 10^{6}\ \text{L/day}$$ $$= 28 \times 20 \times 10^{6}\ \text{mg/day} = 560 \times 10^{6}\ \text{mg/day}$$ $$= 560\ \text{kg/day} \quad (\text{since } 10^{6}\ \text{mg} = 1\ \text{kg})$$ $$\boxed{\text{Alum} = \mathbf{560\ kg/day}}$$

(b) Alkalinity check

Alkalinity consumed by alum:

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

Available raw-water alkalinity $= 12\ \text{mg/L as CaCO}_3$.

Since required (12.6) > available (12), the natural alkalinity is insufficient by:

$$12.6 - 12 = 0.6\ \text{mg/L as CaCO}_3$$

Therefore additional alkalinity must be supplied (as lime). Expressed as CaCO₃ equivalent, the deficit (and hence the make-up alkalinity dose) is:

$$\mathbf{0.6\ mg/L\ as\ CaCO_3}$$

Daily make-up mass:

$$0.6\ \text{mg/L} \times 20 \times 10^{6}\ \text{L/day} = 12 \times 10^{6}\ \text{mg/day} = \mathbf{12\ kg/day\ (as\ CaCO_3)}$$

So a small lime dose equivalent to 0.6 mg/L CaCO₃ (about 12 kg/day) is needed to ensure complete coagulation.

(a) Definitions

• Chlorine demand: the amount of chlorine consumed in oxidising organic matter, ammonia, iron, manganese, etc. It equals applied dose minus residual: $\text{demand} = \text{dose} - \text{residual}$.
• Residual chlorine: the chlorine remaining in water after the demand has been met; it provides continuing disinfection in the distribution system. It may be combined (chloramines) or free (HOCl, OCl⁻).
• Break-point chlorination: dosing chlorine until all demand (especially ammonia) is satisfied and chloramines are destroyed, beyond which any further chlorine appears as free residual.

Break-point curve (described):

Residual
chlorine |        free residual zone
   ^     |               /
   |     |   combined   /
   |     |   residual  /
   |     |    /\      /
   |     |   /  \    /  <- break point (minimum)
   |     |  /    \  /
   |     | /      \/
   +-----+----------------------> Applied chlorine dose

Rising combined residual, then a dip as chloramines are oxidised (the break point), then a linear free-residual rise.

(b) Dose and daily consumption

Required applied dose:

$$\text{Dose} = \text{demand} + \text{desired residual} = 1.6 + 0.5 = \mathbf{2.1\ mg/L}$$

Daily chlorine consumption for $Q = 15\ \text{MLD} = 15 \times 10^{6}\ \text{L/day}$:

$$= 2.1\ \text{mg/L} \times 15 \times 10^{6}\ \text{L/day} = 31.5 \times 10^{6}\ \text{mg/day}$$ $$= \mathbf{31.5\ kg/day}$$

(a) Sources of water

Classification:

• Surface sources: rivers, streams, lakes, ponds, impounding reservoirs.
• Sub-surface / groundwater sources: springs, infiltration galleries, wells (open/dug, tube/bore), aquifers.
• Rainwater (precipitation): roof catchment, rainwater harvesting.

Surface vs groundwater

(b) Quality parameters and guideline values (NDWQS/WHO)

• Turbidity: cloudiness caused by suspended/colloidal matter, measured in NTU. Guideline: 5 NTU (desirable < 1 NTU for effective disinfection).
• Total hardness: sum of multivalent cations (mainly Ca²⁺, Mg²⁺), expressed as mg/L CaCO₃. Guideline: 500 mg/L as CaCO₃ (Nepal NDWQS).
• pH: negative log of H⁺ activity, indicating acidity/alkalinity. Guideline: 6.5–8.5.

(c) Why remove turbidity before disinfection

Suspended/colloidal particles shield micro-organisms from contact with the disinfectant and exert chlorine demand, consuming the disinfectant before it can act on pathogens. Removing turbidity (by coagulation, sedimentation and filtration) ensures the chlorine reaches and inactivates the pathogens efficiently, giving a reliable residual.

Step 1 — Average hourly demand and supply

Daily demand $= 4.5\ \text{ML} = 4{,}500\ \text{m}^3$.

Average hourly demand:

$$= \frac{4{,}500}{24} = 187.5\ \text{m}^3/\text{hr}$$

Supply (uniform pumping) $= 187.5\ \text{m}^3/\text{hr}$ (i.e. 100 % each hour).

Step 2 — Hourly demand by period

Check total demand: $(75+281.25+243.75+150)\times 6 = 750 \times 6 = 4{,}500\ \text{m}^3$ ✓ (= daily demand).

Step 3 — Cumulative storage (running balance)

Starting from 00:00, track cumulative (supply − demand):

Step 4 — Balancing storage

Balancing storage $=$ maximum surplus $+$ maximum deficit (in magnitude):

$$= 675.0 + 225.0 = 900\ \text{m}^3 = \mathbf{0.90\ ML}$$

(Equivalently 20 % of the 4.5 MLD daily demand.)

Step 5 — Total reservoir capacity

$$\text{Total} = \text{balancing} + \text{fire reserve} = 0.90 + 0.40 = \mathbf{1.30\ ML}$$

(A breakdown/emergency reserve could be added in practice; based on the data given the required capacity is 1.30 ML ≈ 1300 m³.)

(a) Components of a gravity-flow scheme

From source to tap:

1. Intake / spring source protection (spring box or stream intake with screen).
2. Collection chamber / sedimentation (silt) tank to remove grit.
3. Transmission/conveyance pipe (HDPE/GI) running by gravity.
4. Break-pressure tanks (BPT) at intervals to limit static head on the pipe.
5. Reservoir / Reservoir-cum-Distribution tank (RVT) for balancing storage.
6. Distribution pipe network.
7. Tapstands (public/private) with drainage, and valve chambers.

(b) Source adequacy check

Present demand (650 people):

$$Q_{demand} = 650 \times 45\ \text{lpcd} = 29{,}250\ \text{L/day}$$

Spring yield per day:

$$Q_{source} = 0.45\ \text{L/s} \times 86{,}400\ \text{s/day} = 38{,}880\ \text{L/day}$$

Since $38{,}880 > 29{,}250$, the source is adequate at present (surplus of 9,630 L/day).

20-year design population (geometric growth, $r = 0.022$):

$$P_{20} = 650 \times (1.022)^{20}$$ $$(1.022)^{20} = e^{20\ln 1.022} = e^{20 \times 0.021761} = e^{0.43522} = 1.5454$$ $$P_{20} = 650 \times 1.5454 = 1{,}004 \approx 1{,}005\ \text{persons}$$

Future demand:

$$= 1{,}005 \times 45 = 45{,}225\ \text{L/day}$$

Compare with dry-season yield 38,880 L/day:

$$45{,}225 > 38{,}880$$

So the spring is NOT adequate for the 20-year design population — it falls short by about 6,345 L/day. An additional/supplementary source, increased storage to ride out peaks, or a reduced service level would be required.

(c) Reasons for community participation

1. Ownership & sustainability — community contribution (labour, local materials, cash) builds a sense of ownership, improving operation, maintenance and tariff collection after handover.
2. Appropriate design & equity — local knowledge ensures the source, tap locations and service level match real needs, and a water-users committee manages fair distribution and upkeep.

(a) Velocity and head loss per km

Diameter $D = 0.30\ \text{m}$; area:

$$A = \frac{\pi}{4}(0.30)^2 = 0.070686\ \text{m}^2$$

Velocity from continuity:

$$V = \frac{Q}{A} = \frac{0.060}{0.070686} = 0.8488\ \text{m/s} \approx \mathbf{0.85\ m/s}$$

Hydraulic radius (full pipe):

$$R = \frac{D}{4} = \frac{0.30}{4} = 0.075\ \text{m}$$

Rearrange Hazen–Williams for $S$:

$$V = 0.849\,C\,R^{0.63}\,S^{0.54} \Rightarrow S = \left(\frac{V}{0.849\,C\,R^{0.63}}\right)^{1/0.54}$$

Compute $R^{0.63}$: $\ln(0.075) = -2.5903$; $\times 0.63 = -1.6319$; $R^{0.63} = e^{-1.6319} = 0.19556$.

Denominator $= 0.849 \times 100 \times 0.19556 = 16.603$.

$$\frac{V}{0.849 C R^{0.63}} = \frac{0.8488}{16.603} = 0.051124$$

Now $S = (0.051124)^{1/0.54}$. Exponent $1/0.54 = 1.85185$.

$$\ln(0.051124) = -2.9737; \quad \times 1.85185 = -5.5069$$ $$S = e^{-5.5069} = 0.004061$$

Head loss per km:

$$h_L = S \times 1000 = 0.004061 \times 1000 = \mathbf{4.06\ m/km}$$

(b) Head loss over 2.4 km and residual head

$$h_{L,2.4} = 4.061 \times 2.4 = 9.75\ \text{m}$$

For a horizontal pipe, residual piezometric (pressure) head downstream:

$$H_{down} = 48 - 9.75 = \mathbf{38.25\ m}$$

So about 9.75 m is lost in friction over 2.4 km, leaving a residual head of roughly 38.3 m at the downstream end — comfortably adequate for distribution.