BE Civil Engineering (IOE, TU) Water Supply Engineering (IOE, CE 651) Question Paper 2080 Nepal

Given census data (decennial, i.e. 10-year intervals). The forecast year 2098 BS is 2 decades (20 years) beyond the last census 2078 BS, so $n = 2$ future decades.

(a) Arithmetic increase method

Decadal increments:

Mean increment per decade:

$$\bar{x} = \frac{6700+8600+10800}{3} = \frac{26100}{3} = 8700$$

Forecast: $P_{2098} = P_{2078} + n\,\bar{x} = 50600 + 2\times 8700 = 50600 + 17400 = \mathbf{68{,}000}$

Geometric increase method

Decadal growth ratios (geometric increase $r$ per decade):

Geometric mean ratio:

$$\bar{R} = (1.27347\times 1.27564\times 1.27136)^{1/3}$$

Product $= 1.27347\times1.27564 = 1.62455$; $\;1.62455\times1.27136 = 2.06532$.

$$\bar{R} = 2.06532^{1/3} = 1.27349$$

so geometric growth rate per decade $r \approx 27.35\%$.

Forecast: $P_{2098} = P_{2078}\,\bar{R}^{\,n} = 50600\times (1.27349)^2$

$$(1.27349)^2 = 1.62178 \Rightarrow P_{2098} = 50600\times 1.62178 = 82{,}062 \approx \mathbf{82{,}060}$$

Summary: Arithmetic ≈ 68,000; Geometric ≈ 82,060. (Geometric is higher because the town is still growing rapidly; arithmetic is more conservative.)

(b) Maximum daily demand (arithmetic population = 68,000)

Average daily demand:

$$Q_{avg} = 68000 \times 135 = 9{,}180{,}000\ \text{L/day} = 9180\ \text{m}^3/\text{day}$$

Maximum daily demand:

$$Q_{max} = 1.8 \times 9180 = \mathbf{16{,}524\ m^3/day}$$

In litres per second:

$$Q_{max} = \frac{16524\times 1000}{86400} = \frac{16{,}524{,}000}{86400} = \mathbf{191.25\ L/s}$$

(a) Stokes' settling velocity

For a particle settling at terminal velocity, the gravitational (submerged) weight = drag force.

Submerged weight: $F_g = \frac{\pi}{6}d^3 (\rho_s-\rho_w)g$

Under laminar flow ($Re<1$) Stokes' drag: $F_D = 3\pi\mu d\,v_s$

Equating $F_g=F_D$:

$$\frac{\pi}{6}d^3(\rho_s-\rho_w)g = 3\pi\mu d\,v_s$$ $$\boxed{v_s = \frac{g(\rho_s-\rho_w)d^2}{18\mu} = \frac{g(S-1)d^2}{18\nu}}$$

Assumptions: particle is spherical, discrete (non-flocculating), settling is laminar ($Re<1$), still fluid, no wall/particle interference, terminal velocity reached.

(b)(i) Settling velocity

$d = 0.02\ \text{mm} = 2\times10^{-5}\ \text{m}$, $S=2.65$, $g=9.81$, $\nu=1.01\times10^{-6}$.

$$v_s = \frac{9.81\,(2.65-1)\,(2\times10^{-5})^2}{18\times 1.01\times10^{-6}}$$

Numerator: $9.81\times1.65 = 16.1865$; $\;(2\times10^{-5})^2 = 4\times10^{-10}$. $16.1865\times4\times10^{-10} = 6.4746\times10^{-9}$.

Denominator: $18\times1.01\times10^{-6} = 1.818\times10^{-5}$.

$$v_s = \frac{6.4746\times10^{-9}}{1.818\times10^{-5}} = 3.561\times10^{-4}\ \text{m/s} = \mathbf{0.356\ mm/s}$$

Check Reynolds number: $Re = \dfrac{v_s d}{\nu} = \dfrac{3.561\times10^{-4}\times2\times10^{-5}}{1.01\times10^{-6}} = 7.05\times10^{-3}$.

Since $Re = 0.007 < 1$, flow is laminar and Stokes' law is valid.

(b)(ii) Surface area and dimensions

For ideal settling, overflow rate (surface loading) = settling velocity:

$$\frac{Q}{A} = v_s \Rightarrow A = \frac{Q}{v_s}$$

$Q = 12\ \text{MLD} = 12{,}000{,}000\ \text{L/day} = 12000\ \text{m}^3/\text{day} = \dfrac{12000}{86400} = 0.1389\ \text{m}^3/\text{s}$.

$$A = \frac{0.1389}{3.561\times10^{-4}} = 390.0\ \text{m}^2$$

With $L = 4B$: $A = L\times B = 4B^2 = 390 \Rightarrow B^2 = 97.5 \Rightarrow B = 9.87\ \text{m}$.

$L = 4\times9.87 = 39.5\ \text{m}$.

Provide tank ≈ 40 m (L) × 10 m (B) (surface area ≈ 400 m², slightly conservative). Final answer: A ≈ 390 m², dimensions ≈ 40 m × 10 m.

(a) Rapid Sand Filter

  Inlet (treated/coagulated water)
        |
   +----v-------------------------+  <- wash-water trough
   |  ~~~~~ water over sand ~~~~~  |
   |  ::::::: SAND BED :::::::::   |  (0.6-0.75 m, effective size ~0.45-0.7 mm)
   |  ooooo GRAVEL LAYERS ooooo   |  (graded, ~0.45 m)
   |  ___ under-drain laterals ___|--> filtered water -> clear well
   +------------------------------+
        ^ backwash water in (during cleaning)

Working: Coagulated, flocculated and settled water enters above the sand bed and percolates downward. Suspended impurities and floc are removed mainly by straining, sedimentation within pores, and adsorption in the top few cm of the sand. Filtered water is collected by the under-drainage system and led to the clear-water reservoir, then disinfected.

Backwashing: As filtration proceeds the head loss rises (clogging). When terminal head loss is reached the filter is cleaned by reversing the flow — clean water is forced upward through the under-drains at high rate, expanding (fluidising) the sand bed so that trapped dirt is scoured and carried away through wash-water troughs. Air scour may assist. Backwashing restores porosity/permeability and is essential to keep the filtration rate and water quality.

(b)(i) Filter area & number of units

$Q = 24\ \text{MLD} = 24{,}000{,}000\ \text{L/day}$.

Filtration rate $= 5000\ \text{L/m}^2/\text{h} = 5000\times24 = 120{,}000\ \text{L/m}^2/\text{day}$.

Total area:

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

Number of working units (each ≤ 40 m²): $200/40 = 5$ units. Add 1 stand-by ⇒ 6 units of 40 m² each (5 working + 1 stand-by).

(b)(ii) Backwash water loss

Wash-water volume per unit per wash:

$$V_{bw} = 0.4\ \tfrac{\text{m}^3}{\text{m}^2\cdot\text{min}}\times 40\ \text{m}^2 \times 8\ \text{min} = 128\ \text{m}^3$$

With 5 working units washed once/day: total $= 5\times128 = 640\ \text{m}^3/\text{day}$.

Treated water $= 24\ \text{MLD} = 24{,}000\ \text{m}^3/\text{day}$.

$$\%\ \text{loss} = \frac{640}{24000}\times100 = \mathbf{2.67\%}$$

(a) Dead-end vs Grid-iron

(b) Hazen–Williams head loss

$L=600\ \text{m}$, $Q=45\ \text{L/s}=0.045\ \text{m}^3/\text{s}$, $D=0.30\ \text{m}$, $C=120$.

$$h_f = \frac{10.67\,Q^{1.852}\,L}{C^{1.852}\,D^{4.87}}$$

Compute each term:

$Q^{1.852}$: $\ln(0.045)=-3.10109$; $\times1.852=-5.74322$; $e^{-5.74322}=3.2099\times10^{-3}$.

$C^{1.852}=120^{1.852}$: $\ln120=4.78749$; $\times1.852=8.86643$; $e^{8.86643}=7106.5$.

$D^{4.87}=0.30^{4.87}$: $\ln0.30=-1.20397$; $\times4.87=-5.86335$; $e^{-5.86335}=2.8467\times10^{-3}$.

Numerator: $10.67\times3.2099\times10^{-3}\times600 = 10.67\times1.92594 = 20.550$.

Denominator: $7106.5\times2.8467\times10^{-3} = 20.231$.

$$h_f = \frac{20.550}{20.231} = \mathbf{1.016\ m} \approx 1.02\ \text{m}$$

(ii) Residual head

Pipe horizontal, so elevation head unchanged. Residual pressure head at downstream end:

$$H_{res} = H_{available} - h_f = 40 - 1.016 = \mathbf{38.98\ m} \approx 39.0\ \text{m of water}$$

This comfortably exceeds the minimum residual pressure (typically 10–17 m), so the supply is adequate.

(a) Total manometric head

$Q = 40\ \text{L/s} = 0.04\ \text{m}^3/\text{s}$, $D=0.25\ \text{m}$.

Area $A = \dfrac{\pi}{4}D^2 = 0.7854\times0.0625 = 0.049087\ \text{m}^2$.

Velocity $V = \dfrac{Q}{A} = \dfrac{0.04}{0.049087} = 0.8149\ \text{m/s}$.

Friction head:

$$h_f = \frac{fLV^2}{2gD} = \frac{0.02\times220\times(0.8149)^2}{2\times9.81\times0.25}$$

$V^2 = 0.66406$. Numerator $= 0.02\times220\times0.66406 = 2.92186$. Denominator $= 4.905$.

$$h_f = \frac{2.92186}{4.905} = 0.5957\ \text{m}$$

Minor losses $= 0.15\times0.5957 = 0.0894\ \text{m}$.

Total head:

$$H = H_{static} + h_f + h_{minor} = 28 + 0.5957 + 0.0894 = \mathbf{28.69\ m}$$

(b) Power and energy

Water (output) power:

$$P_{water} = \rho g Q H = 1000\times9.81\times0.04\times28.69 = 11{,}258\ \text{W} = 11.26\ \text{kW}$$

Input (shaft/electrical) power at 72% overall efficiency:

$$P_{input} = \frac{P_{water}}{\eta} = \frac{11.26}{0.72} = \mathbf{15.64\ kW}$$

Daily energy at 10 h/day:

$$E = P_{input}\times t = 15.64\times10 = \mathbf{156.4\ kWh/day}$$

(a) Coagulation & flocculation

Colloidal turbidity particles carry negative surface charges and repel one another, so they stay suspended. A coagulant (e.g. alum) added and rapidly mixed releases trivalent $\text{Al}^{3+}$ ions that neutralise these charges (compress the electric double layer) and form gelatinous aluminium hydroxide $\text{Al(OH)}_3$ floc. Coagulation is this rapid destabilisation/charge-neutralisation; flocculation is the subsequent slow, gentle stirring that lets destabilised particles collide and agglomerate into larger, heavier flocs that settle readily in the sedimentation tank.

Alum is most common because it is cheap, readily available, easy to handle/store, effective over a useful pH range (~5.5–8), and produces a dense, fast-settling floc.

(b)(i) Alum required per day

$Q = 8\ \text{MLD} = 8\times10^{6}\ \text{L/day}$. Dose $=22\ \text{mg/L}$.

$$\text{Mass} = 8\times10^{6}\ \tfrac{\text{L}}{\text{day}}\times 22\ \tfrac{\text{mg}}{\text{L}} = 1.76\times10^{8}\ \text{mg/day} = \mathbf{176\ kg/day}$$

(Using $1\ \text{mg/L per MLD} = 1\ \text{kg/day}$: $8\times22 = 176\ \text{kg/day}$.)

(b)(ii) Alkalinity consumed

Alkalinity consumed $= 0.5\times \text{alum dose} = 0.5\times22 = 11\ \text{mg/L as CaCO}_3$.

Mass per day $= 8\times11 = \mathbf{88\ kg/day\ as\ CaCO_3}$.

(a) Break-point chlorination

Break-point chlorination is the application of chlorine just beyond the point at which the chlorine demand of the water is fully satisfied, so that any further chlorine added appears as free available residual chlorine.

Typical curve (residual chlorine on y-axis vs applied dose on x-axis):

Residual
  |              /
  |    /\       /  <- free residual (slope ~45°)
  |   /  \     /
  |  /    \   /  <- break point (minimum)
  | /      \ /
  |/________V________ Dose
   demand  combined  break

• First, chlorine reacts with reducing agents (no residual).
• Then it forms chloramines (combined residual) with ammonia — residual rises.
• Continued dosing oxidises/destroys chloramines — residual falls to a minimum: the break point.
• Beyond it, residual rises again as free chlorine.

Significance: dosing past the break point guarantees a reliable disinfecting free residual, destroys taste/odour-causing chloramines and oxidises ammonia/organics.

(b)(i) Chlorine dose & daily requirement

Dose $=$ demand $+$ residual $= 2.4 + 0.5 = 2.9\ \text{mg/L}$.

$Q = 6\ \text{MLD}$. Daily chlorine $= 6\times2.9 = \mathbf{17.4\ kg/day}$.

(check: $6\times10^6\ \text{L/day}\times2.9\ \text{mg/L}=1.74\times10^7\ \text{mg/day}=17.4\ \text{kg/day}$.)

(b)(ii) Bleaching powder (30% available chlorine)

$$\text{Bleaching powder} = \frac{\text{chlorine required}}{0.30} = \frac{17.4}{0.30} = \mathbf{58\ kg/day}$$

(a) Sources of water

Surface sources — rivers/streams, lakes & ponds, impounding reservoirs, and runoff stored from rainfall.

Sub-surface (ground) sources — springs, infiltration galleries, open (dug) wells, and tube/bore wells; also rainwater (harvesting) as a supplementary source.

Groundwater (springs/wells) in Nepal's hills:

• Merit: generally of good bacteriological quality and low turbidity, so it needs little or no treatment (often only disinfection); flow can be relatively steady through dry months.
• Demerit: yield is often limited/seasonal, sources are dispersed and at varying elevations making collection and gravity conveyance difficult; spring discharge can drop sharply in the dry season.

(b) Intake structures

An intake structure is a device/structure constructed in a surface source (river, lake, reservoir) to draw (admit) water from the source and convey it to the treatment works / pumping station, while keeping out floating matter, debris and excessive silt.

Requirements of a good intake:

• Located where the purest, most reliable water is available in all seasons.
• Should draw water even at lowest water level; safe in floods.
• Protected from silt, debris and floating matter (screens).
• Structurally stable against currents, scour and ice; easy to operate and maintain.

Types of river intakes: submerged intake and exposed (intake tower) intake; broadly classified as wet intake and dry intake towers; also canal intake for diversion schemes.

(a) Quality parameters

Physical parameters are judged by senses/instruments; chemical by titration/spectrometry; bacteriological by culture (MPN / membrane filter) as indicators of faecal contamination.

(b) Potability comment

Observed MPN = 16 coliforms/100 mL, whereas the standard for treated drinking water is 0/100 mL.

Since coliforms are present (and well above the limit of zero), the water is bacteriologically unsafe / non-potable — it indicates faecal/sewage contamination and risk of water-borne disease (diarrhoea, cholera, typhoid).

Corrective measures:

• Disinfect the water (chlorination to a free residual of ~0.5 mg/L, boiling, or SODIS/UV at household level).
• Identify and eliminate the contamination source (protect intake/reservoir, repair leaks/cross-connections, fence and protect the spring/well head).
• Re-test after treatment to confirm 0 coliforms/100 mL before supply.

Step 1 — Maximum daily demand

$$Q_{max} = 1.5\times Q_{avg} = 1.5\times 5400 = 8100\ \text{m}^3/\text{day}$$

Step 2 — Balancing (equalising) storage

$$V_{bal} = 0.35\times Q_{max} = 0.35\times 8100 = 2835\ \text{m}^3$$

Step 3 — Breakdown / emergency storage

$$V_{bd} = \tfrac{1}{3}\times Q_{avg} = \tfrac{1}{3}\times 5400 = 1800\ \text{m}^3$$

Step 4 — Fire storage

$$V_{fire} = 100\ \text{m}^3$$

Step 5 — Total capacity

$$V_{total} = V_{bal}+V_{bd}+V_{fire} = 2835 + 1800 + 100 = \mathbf{4735\ m^3}$$

Required service-reservoir capacity ≈ 4735 m³ (provide ~4750 m³).

Note: the 16 h/day uniform pumping (vs. 24 h variable demand) is what creates the need for the balancing storage component; here it is taken empirically as 35% of $Q_{max}$.

(a) Gravity-flow rural scheme

Salient features: water flows entirely under gravity from a higher source to lower settlements (no pumping/energy cost); suited to Nepal's hill topography; uses a protected spring/stream source; low O&M; commonly community-managed.

Main components (source → tap): Source works / intake (spring source protection) → sedimentation/collection (reservoir/intake chamber) → conveyance (transmission) main → break-pressure tanks (to limit static head in steep terrain) → reservoir / service tank → distribution pipeline → public tap stands / household taps.

(b) Head check

Available (static) head $= \text{RL}_{source} - \text{RL}_{tank} = 1240 - 1180 = 60\ \text{m}$.

Head loss along pipe $= 0.05\ \text{m/m}\times 900\ \text{m} = 45\ \text{m}$.

Since required head loss $45\ \text{m} < 60\ \text{m}$ available, the head is sufficient to convey the design flow.

Residual (surplus) head at tank:

$$H_{res} = 60 - 45 = \mathbf{15\ m}$$

The 15 m surplus should be dissipated by a break-pressure tank (or absorbed by the tank's static head) so that pipe and fittings are not over-pressurised.