BE Civil Engineering (IOE, TU) Water Supply Engineering (IOE, CE 651) Question Paper 2079 Nepal

(a) Population projection to 2041

The last census is 2021, so the design year 2041 is 2 decades ahead ($n = 2$).

Decadal increases:

Arithmetic increase method

Average decadal increase:

$$\bar{x} = \frac{6{,}000 + 8{,}000 + 10{,}000 + 14{,}000}{4} = \frac{38{,}000}{4} = 9{,}500$$ $$P_{2041} = P_{2021} + n\,\bar{x} = 62{,}000 + 2(9{,}500) = 62{,}000 + 19{,}000 = \mathbf{81{,}000}$$

Geometric increase method

Decadal growth rates:

Geometric mean of growth ratios $(1+r)$:

$$1+r = \sqrt[4]{(1.2500)(1.2667)(1.2632)(1.2917)}$$

Product $= 1.2500 \times 1.2667 = 1.58338$; $\;1.58338 \times 1.2632 = 2.00014$; $\;2.00014 \times 1.2917 = 2.58358$.

$$1+r = (2.58358)^{1/4} = 1.2680 \;\Rightarrow\; r = 0.2680\ (26.80\%\ \text{per decade})$$ $$P_{2041} = P_{2021}(1+r)^n = 62{,}000\,(1.2680)^2 = 62{,}000 \times 1.60782 = \mathbf{99{,}685 \approx 99{,}700}$$

(b) Maximum daily demand for 2041 (geometric population)

Use $P_{2041} = 99{,}685$, demand $= 135$ lpcd, peak factor $= 1.8$.

Average daily demand:

$$Q_{avg} = \frac{99{,}685 \times 135}{1000} = 13{,}457.5\ \text{m}^3/\text{day}$$

Maximum daily demand:

$$Q_{max} = 1.8 \times 13{,}457.5 = 24{,}223.5\ \text{m}^3/\text{day}$$

Convert to litres per second:

$$Q_{max} = \frac{24{,}223.5 \times 1000}{86{,}400} = 280.4\ \text{L/s}$$

Maximum daily demand ≈ 24,224 m³/day ≈ 280.4 L/s.

Given

$Q = 18\ \text{MLD} = 18{,}000\ \text{m}^3/\text{day}$; overflow rate $v_0 = 30\ \text{m}^3/\text{m}^2/\text{day}$; detention $t = 3\ \text{h}$.

(a) Surface area and plan dimensions

$$A = \frac{Q}{v_0} = \frac{18{,}000}{30} = 600\ \text{m}^2$$

With $L = 4B$: $\;A = L\times B = 4B^2 = 600 \Rightarrow B^2 = 150 \Rightarrow B = 12.25\ \text{m}$.

$$L = 4 \times 12.25 = 49.0\ \text{m}$$

Adopt L = 49.0 m, B = 12.25 m (area = 600.25 m² ≈ 600 m²).

(b) Depth, volume and detention check

Volume required for 3 h detention:

$$V = Q \times t = 18{,}000\ \text{m}^3/\text{day} \times \frac{3}{24}\ \text{day} = 2{,}250\ \text{m}^3$$

Depth:

$$H = \frac{V}{A} = \frac{2{,}250}{600} = 3.75\ \text{m}$$

Check detention time: $t = V/Q = 2{,}250 / 18{,}000 = 0.125\ \text{day} = 3.0\ \text{h}$ ✓

Adopt depth = 3.75 m (plus free board ≈ 0.5 m).

(c) Stokes' law settling velocity

$d = 0.02\ \text{mm} = 2 \times 10^{-5}\ \text{m}$, $G_s = 2.65$, $\mu = 1.005\times 10^{-3}\ \text{N·s/m}^2$.

$$v_s = \frac{g\,(G_s - 1)\,d^2}{18\,\nu}\quad\text{(using kinematic form)} = \frac{g\,(\rho_s-\rho_w)\,d^2}{18\,\mu}$$ $$v_s = \frac{9.81 \times (2650-1000) \times (2\times10^{-5})^2}{18 \times 1.005\times10^{-3}}$$

Numerator $= 9.81 \times 1650 \times 4\times10^{-10} = 6.4746\times10^{-6}$. Denominator $= 0.01809$.

$$v_s = \frac{6.4746\times10^{-6}}{0.01809} = 3.578\times10^{-4}\ \text{m/s}$$

Convert overflow rate to m/s: $v_0 = 30\ \text{m/day} = 30/86400 = 3.472\times10^{-4}\ \text{m/s}$.

Since $v_s = 3.58\times10^{-4}\ \text{m/s} > v_0 = 3.47\times10^{-4}\ \text{m/s}$, the particle settling velocity exceeds the overflow rate, so the particle WILL be fully removed in the tank.

(Check Reynolds number: $Re = v_s d/\nu = 3.578\times10^{-4}\times2\times10^{-5}/(1.005\times10^{-6}) = 0.0071 < 1$, so Stokes' law is valid.)

Given

$Q = 24\ \text{MLD} = 24{,}000{,}000\ \text{L/day}$; filtration rate $= 5000\ \text{L/m}^2/\text{h}$.

(a) Filter area and number of units

Daily filtration capacity per m²: $5000\ \text{L/m}^2/\text{h} \times 24\ \text{h} = 120{,}000\ \text{L/m}^2/\text{day} = 120\ \text{m}^3/\text{m}^2/\text{day}$.

Total area:

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

Number of working units at 40 m² each: $200/40 = 5$ units.

Provide 1 stand-by → total 6 units. (When one unit is washed/down, the remaining 5 still deliver design flow.)

Actual area per working unit when all 6 are in service can be checked, but design adopts 6 units of 40 m² each (5 working + 1 stand-by).

(b) Wash-water as percentage of output

Area of one unit $= 40\ \text{m}^2$. Wash rate $= 0.5\ \text{m}^3/\text{m}^2/\text{min}$ for $8\ \text{min}$.

Wash water per unit per wash:

$$= 0.5 \times 40 \times 8 = 160\ \text{m}^3$$

There are 5 working units (each washed once/day); total daily wash water:

$$= 160 \times 5 = 800\ \text{m}^3/\text{day}$$

Daily filtered output $= 24{,}000\ \text{m}^3/\text{day}$.

$$\%\text{ wash water} = \frac{800}{24{,}000}\times100 = 3.33\%$$

Wash water ≈ 3.33% of filtered output (within the acceptable 2–6% range).

(c) Functions

Gravel support layer:

1. Supports the fine sand bed and prevents fine media from passing into the under-drains.
2. Distributes the rising backwash water uniformly across the filter bottom.

Under-drainage system:

1. Collects the filtered water and conveys it to the clear-water well.
2. Distributes backwash water (and air, if used) evenly during cleaning of the bed.

Negative head: As the filter run progresses the sand bed clogs and head loss through the upper layers increases. If the head loss exceeds the depth of water standing above a given level in the bed, the pressure there falls below atmospheric (a partial vacuum), called negative head. It causes release of dissolved gases (air binding) which further blocks the pores and reduces the filtration rate; backwashing is then required.

Hardy-Cross correction formula

For a loop with $h_f = KQ^2$:

$$\Delta = -\frac{\sum K Q |Q|}{\sum 2K|Q|}$$

Taking clockwise as positive. Initial assumed flows satisfy continuity:

• Node A: in 120, out to AB (70) and from DA. Check node B: in 70, out 40, to BC 30 ✓. Node C: in 30, out 30, CD 0 ✓. Node D: CD in 0, DA out 50, demand 50 ✓.

Iteration 1 – build the table

| Pipe | K | Q (L/s) | $KQ|Q|$ | $2K|Q|$ | |------|------|---------|---------|---------| | AB | 0.0020 | +70 | 0.0020·70·70 = +9.800 | 2·0.0020·70 = 0.280 | | BC | 0.0050 | +30 | 0.0050·30·30 = +4.500 | 2·0.0050·30 = 0.300 | | CD | 0.0030 | 0 | 0.000 | 0.000 | | DA | 0.0010 | −50 | 0.0010·(−50)·50 = −2.500 | 2·0.0010·50 = 0.100 |

$$\sum KQ|Q| = 9.800 + 4.500 + 0.000 - 2.500 = +11.800$$ $$\sum 2K|Q| = 0.280 + 0.300 + 0.000 + 0.100 = 0.680$$

Correction

$$\Delta = -\frac{11.800}{0.680} = -17.35\ \text{L/s}$$

Apply $\Delta$ to every pipe of the loop (add algebraically, keeping sign convention):

Corrected flows after one iteration

• $Q_{AB} = 52.65\ \text{L/s}$ (A→B)
• $Q_{BC} = 12.65\ \text{L/s}$ (B→C)
• $Q_{CD} = 17.35\ \text{L/s}$ (C→D direction reversed to D→C, i.e. flow actually D→C)
• $Q_{DA} = 67.35\ \text{L/s}$ (flow A→D since sign negative w.r.t. clockwise)

Continuity re-check at B: in 52.65, out demand 40 + BC 12.65 = 52.65 ✓.

A further iteration would be performed until $|\Delta| \le$ allowable (≈ 0.1–1 L/s); after one iteration $\Delta = -17.35$ L/s is still large, so iteration continues in practice.

(b) Head loss in pipe AB after first iteration

Using corrected $Q_{AB} = 52.65\ \text{L/s}$ and $K_{AB} = 0.0020$:

$$h_{f,AB} = K Q^2 = 0.0020 \times (52.65)^2 = 0.0020 \times 2772.0 = 5.544\ \text{(head-loss units)}$$

Head loss in AB ≈ 5.54 m (in the consistent units of $K$).

Given

$Q = 40\ \text{L/s} = 0.040\ \text{m}^3/\text{s}$; static lift $H_s = 35\ \text{m}$; $D = 0.250\ \text{m}$; $L = 480\ \text{m}$; $f = 0.024$.

(a) Total dynamic head

Velocity in rising main:

$$A = \frac{\pi}{4}D^2 = \frac{\pi}{4}(0.250)^2 = 0.04909\ \text{m}^2$$ $$V = \frac{Q}{A} = \frac{0.040}{0.04909} = 0.8149\ \text{m/s}$$

Friction loss (Darcy–Weisbach):

$$h_f = \frac{f L V^2}{2 g D} = \frac{0.024 \times 480 \times (0.8149)^2}{2 \times 9.81 \times 0.250}$$

Numerator $= 0.024 \times 480 \times 0.66406 = 7.6516$. Denominator $= 4.905$.

$$h_f = \frac{7.6516}{4.905} = 1.560\ \text{m}$$

Minor losses $= 0.15 \times h_f = 0.15 \times 1.560 = 0.234\ \text{m}$.

Velocity head (delivered) $= V^2/2g = 0.66406/19.62 = 0.034\ \text{m}$ (small, may be included).

Total dynamic head:

$$H = H_s + h_f + h_{minor} + \frac{V^2}{2g} = 35 + 1.560 + 0.234 + 0.034 = 36.83\ \text{m}$$

Total dynamic (manometric) head H ≈ 36.83 m (≈ 36.8 m if velocity head neglected: 36.79 m).

(b) Power input and daily energy

Water (output) power:

$$P_{water} = \rho g Q H = 1000 \times 9.81 \times 0.040 \times 36.83 = 14{,}452\ \text{W} = 14.45\ \text{kW}$$

Input power at overall efficiency $\eta = 0.72$:

$$P_{input} = \frac{P_{water}}{\eta} = \frac{14.45}{0.72} = 20.07\ \text{kW}$$

Daily energy (16 h/day):

$$E = 20.07\ \text{kW} \times 16\ \text{h} = 321.1\ \text{kWh/day}$$

Power input ≈ 20.07 kW; daily energy ≈ 321 kWh/day.

(c) Water hammer

Water hammer is the sudden rise (and subsequent oscillation) of pressure in a pipeline caused by a rapid change in the velocity of flow — for example when a valve is closed quickly or a pump trips. The momentum of the moving water column is abruptly arrested, producing a pressure surge that can burst pipes or damage fittings.

Two control devices: (i) Air vessel / surge tank (absorbs and cushions the pressure surge); (ii) Slow-closing or automatic relief/non-return valves (limit the rate of velocity change and release excess pressure). Flywheels on pump motors are also used to slow the velocity change.

(a) Chlorine mass and residual

$Q = 6\ \text{MLD} = 6 \times 10^{6}\ \text{L/day} = 6{,}000\ \text{m}^3/\text{day}$; dose $= 2.5\ \text{mg/L}$.

Mass of chlorine per day:

$$M = \text{dose} \times Q = 2.5\ \frac{\text{mg}}{\text{L}} \times 6\times10^{6}\ \frac{\text{L}}{\text{day}} = 15 \times 10^{6}\ \text{mg/day}$$ $$= 15{,}000{,}000\ \text{mg/day} = 15\ \text{kg/day}$$

Residual chlorine $=$ dose $-$ demand $= 2.5 - 1.8 = 0.7\ \text{mg/L}$.

Chlorine required = 15 kg/day; residual maintained = 0.7 mg/L (within the recommended 0.2–0.5 mg/L free residual range at the consumer end after travel; at the plant 0.7 mg/L is acceptable).

(b) Break-point chlorination

Break-point chlorination is the application of chlorine in a dose large enough to satisfy the entire chlorine demand — oxidising organics, ammonia and reducing compounds and destroying chloramines — so that any chlorine added beyond the break point appears as free available chlorine residual.

A free chlorine residual is desirable because:

• It provides a continued disinfecting (residual) capacity throughout the distribution system, guarding against re-contamination.
• Free chlorine is a stronger and faster disinfectant than combined (chloramine) residual and serves as an indicator that disinfection is adequate.

(a) Daily alum requirement

$Q = 10\ \text{MLD} = 10 \times 10^{6}\ \text{L/day}$; dose $= 30\ \text{mg/L}$.

$$M_{alum} = 30\ \frac{\text{mg}}{\text{L}} \times 10\times10^{6}\ \frac{\text{L}}{\text{day}} = 300\times10^{6}\ \text{mg/day} = 300\ \text{kg/day}$$

Alum requirement = 300 kg/day.

(b) Alkalinity check and lime dose

Alkalinity required to react with alum:

$$= 0.45 \times \text{alum dose} = 0.45 \times 30 = 13.5\ \text{mg/L (as CaCO}_3)$$

Natural alkalinity available $= 18\ \text{mg/L}$.

Since $18\ \text{mg/L} > 13.5\ \text{mg/L}$, the natural alkalinity IS sufficient; there is a surplus of $18 - 13.5 = 4.5\ \text{mg/L}$ as CaCO₃.

No quicklime addition is required for coagulation.

(If alkalinity had been deficient, the lime dose would be CaO $= 0.35 \times$ deficit. Here deficit $= 0$, so CaO $= 0$ mg/L.)

(a) Surface vs groundwater sources

(Any two valid points on quality and on reliability earn full marks.)

(b) Drinking-water parameters and NDWQS limits

Four parameters: turbidity, pH, total hardness (as CaCO₃), and total/faecal coliform (bacteriological). Others: arsenic, iron, chloride, nitrate.

Maximum permissible values per NDWQS 2005 (examples):

• Turbidity: 5 NTU (max permissible 10 NTU)
• pH: 6.5 – 8.5
• Total hardness (as CaCO₃): 500 mg/L
• Arsenic: 0.05 mg/L
• E. coli / total coliform: 0 per 100 mL (must be absent)

(Stating any two correctly is sufficient for the 2 marks.)

(a) Average daily demand

$$Q = 600\ \text{persons} \times 45\ \text{lpcd} = 27{,}000\ \text{L/day}$$

Average daily demand = 27,000 L/day (= 27 m³/day).

(b) Reservoir storage capacity

Storage $= 50\%$ of average daily demand:

$$V = 0.50 \times 27{,}000 = 13{,}500\ \text{L} = 13.5\ \text{m}^3$$

Required storage capacity = 13.5 m³.

(c) Source flow if collected over 12 hours

$$q = \frac{27{,}000\ \text{L/day}}{12\ \text{h} \times 60\ \text{min/h}} = \frac{27{,}000}{720} = 37.5\ \text{L/min}$$

Required source supply rate = 37.5 L/min (≈ 0.625 L/s).

(d) Reason for distribution (service) storage

It balances the variation between the steady source/supply inflow and the fluctuating hourly demand of the village — storing water during low-demand (night) hours and meeting peak (morning/evening) draw-off, while also providing a reserve during source interruption or maintenance.

(a) Distribution-system layouts

1. Dead-end (tree) system

        main
  o-----+-----+-----+-----o
        |     |     |
        +     +     +   (branch sub-mains, dead ends)

Water flows from one main to progressively smaller branches that terminate in dead ends. Advantage: Simple to design and lay; fewer pipes and valves, hence economical and easy to compute discharges.

2. Grid-iron (interconnected) system

  +----+----+----+
  |    |    |    |
  +----+----+----+
  |    |    |    |
  +----+----+----+

Mains and sub-mains are interconnected so water can reach any point by more than one path. Advantage: No dead ends — water remains fresh (less stagnation), and during a repair flow is maintained from alternate routes; better pressure distribution.

(Ring system: a closed loop main surrounding the area — advantage: any node fed from two directions. Radial system: water pumped outward from central reservoirs — advantage: high pressure and quick service.)

(b) Valves / appurtenances

• Sluice (gate) valve: isolates a pipe section for repair by stopping flow.
• Air-relief valve: releases accumulated air at summits to prevent air locks and pressure problems.
• Scour (washout) valve: placed at low points to flush out accumulated silt/sediment from the mains; a check (reflux/non-return) valve prevents reverse flow on a pumping main.

(a) Average daily domestic demand

$$Q_{dom} = 150{,}000 \times 150\ \text{lpcd} = 22{,}500{,}000\ \text{L/day} = 22{,}500\ \text{m}^3/\text{day}$$

Average daily domestic demand = 22,500 m³/day.

(b) Fire demand (Freeman-type formula)

$P = 150$ (population in thousands), $\sqrt{P} = \sqrt{150} = 12.247$.

$$Q = 4637\sqrt{P}\,(1 - 0.01\sqrt{P}) = 4637 \times 12.247 \times (1 - 0.01\times12.247)$$

$= 4637 \times 12.247 \times (1 - 0.12247)$ $= 4637 \times 12.247 \times 0.87753$

$4637 \times 12.247 = 56{,}789$ (L/min, before factor). $56{,}789 \times 0.87753 = 49{,}833\ \text{L/min}$.

Convert to daily volume: $49{,}833\ \text{L/min} \times 1440\ \text{min/day} = 71.76 \times 10^{6}\ \text{L/day} \approx 71{,}760\ \text{m}^3/\text{day}$ (if sustained 24 h; usually applied for a few hours).

Fire demand ≈ 49,833 L/min ≈ 830 L/s.

(c) Coincident-draft principle

Fire demand is not added to the average demand. It is provided in addition to the maximum daily demand, because a fire is assumed to occur at a time of high (maximum daily) consumption. Thus the design draft (for source/pump/main capacity) is taken as the greater of (maximum hourly demand) and (maximum daily demand + fire demand).