BE Civil Engineering (IOE, TU) Water Supply Engineering (IOE, CE 651) Question Paper 2077 Nepal

Decadal data and increments

(a) Geometric Increase method

Geometric mean of the percentage increases (use the decadal growth ratios):

$$r_g = \sqrt[3]{0.2500 \times 0.3000 \times 0.28205} = \sqrt[3]{0.021154} = 0.27645 \;\Rightarrow\; 27.645\% \text{ per decade}$$

Forecast to 2041 = 2 decades beyond 2021:

$$P_n = P_0 (1+r_g)^n = 50000 \times (1.27645)^2$$ $$= 50000 \times 1.62933 = 81{,}466 \approx \mathbf{81{,}500 \text{ persons}}$$

Incremental Increase method

Average increase $\bar{x} = \dfrac{6000+9000+11000}{3} = \dfrac{26000}{3} = 8{,}666.7$

Average incremental increase $\bar{y} = \dfrac{3000+2000}{2} = 2{,}500$

$$P_n = P_0 + n\bar{x} + \frac{n(n+1)}{2}\bar{y}, \quad n = 2$$ $$= 50000 + 2(8666.7) + \frac{2 \times 3}{2}(2500) = 50000 + 17333.4 + 7500 = \mathbf{74{,}833 \approx 74{,}800 \text{ persons}}$$

(b) Maximum daily demand (geometric forecast = 81,466)

Average daily demand:

$$Q_{avg} = \frac{81466 \times 135}{1000} = 10{,}998 \;\text{m}^3/\text{day}$$

Maximum daily demand:

$$Q_{max} = 1.8 \times 10998 = \mathbf{19{,}797 \approx 19{,}800 \;\text{m}^3/\text{day}}$$

Limitation: The geometric method assumes a constant percentage growth and tends to over-estimate for towns approaching saturation. Nepali hill towns often grow by migration spurts and out-migration, so growth is irregular; the geometric method ignores topographic and economic ceilings on growth.

(a) Settling velocity (Stokes' law)

For a particle of diameter $d$, density $\rho_s$ settling in water of density $\rho_w$ and dynamic viscosity $\mu$:

Net downward force (gravity − buoyancy):

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

Drag force in laminar flow (Stokes): $F_D = 3\pi \mu d \, v_s$

At terminal (settling) velocity, $F_g = F_D$:

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

This is valid for Reynolds number $R_e = \rho_w v_s d/\mu < 1$.

(b) Sedimentation tank design

Flow: $Q = 6000\,\text{m}^3/\text{day} = \dfrac{6000}{86400} = 0.06944\,\text{m}^3/\text{s}$

Surface area (overflow rate = settling velocity of design particle):

$$A = \frac{Q}{v_s} = \frac{0.06944}{0.0005} = 138.9\,\text{m}^2 \approx \mathbf{139\,\text{m}^2}$$

Dimensions with $L = 4B$:

$$A = L \times B = 4B^2 = 138.9 \Rightarrow B^2 = 34.72 \Rightarrow B = 5.89\,\text{m}$$

Adopt $B = 6\,\text{m}$, $L = 24\,\text{m}$ (provided area $= 144\,\text{m}^2 \ge 139$, OK).

Volume and detention time (depth = 3 m):

$$V = L \times B \times H = 24 \times 6 \times 3 = 432\,\text{m}^3$$ $$t_d = \frac{V}{Q} = \frac{432}{0.06944} = 6221\,\text{s} = \mathbf{1.73\,\text{hours}}$$

Horizontal flow velocity: Cross-sectional area $= B \times H = 6 \times 3 = 18\,\text{m}^2$

$$v_h = \frac{Q}{B \times H} = \frac{0.06944}{18} = 0.00386\,\text{m/s} = \mathbf{3.86\,\text{mm/s}}$$

Comment: Horizontal velocity of 3.86 mm/s is well below the usual limit of ~9 mm/s (and below the scour velocity), so no re-suspension of settled flocs is expected. Detention time of ~1.7 h is slightly low for plain sedimentation (typical 2–4 h); depth or area could be increased if a longer detention is desired.

(a) Slow vs Rapid sand filter

(b) Rapid sand filter sizing

Flow: $Q = 4\,\text{MLD} = 4{,}000{,}000\,\text{L/day}$

Filtration over 24 h, rate $= 5000\,\text{L/m}^2/\text{h}$:

$$\text{Hourly flow} = \frac{4{,}000{,}000}{24} = 166{,}667\,\text{L/h}$$ $$\text{Required area} = \frac{166{,}667}{5000} = 33.33\,\text{m}^2$$

Area of one bed $= 5 \times 4 = 20\,\text{m}^2$

Number of working beds $= \dfrac{33.33}{20} = 1.67 \Rightarrow$ 2 working beds

Provide 1 standby $\Rightarrow$ total = 3 beds.

Provided working area $= 2 \times 20 = 40\,\text{m}^2$.

Filtration rate when standby in service (i.e. one working bed out, all 3 used = still only need to cover Q with available beds; with 2 beds working the rate is):

$$\text{Rate per working bed (2 beds)} = \frac{166{,}667}{2 \times 20} = \mathbf{4{,}167\,\text{L/m}^2/\text{h}}$$

If one of the three beds is under backwash and only 2 remain in service, the rate is 4,167 L/m²/h, which is below the design maximum of 5,000 L/m²/h — so operation with the standby maintains a safe filtration rate. Adopt 3 beds, each 5 m × 4 m.

Given data

• $Q = 40\,\text{L/s} = 0.040\,\text{m}^3/\text{s}$, $L = 1500\,\text{m}$, $C = 120$
• Available static head $= 100.0 - 70.0 = 30.0\,\text{m}$
• Required residual head at B $\ge 12\,\text{m}$ $\Rightarrow$ allowable head loss $h_{f,allow} = 30 - 12 = 18\,\text{m}$

Hazen–Williams constant part:

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

$Q^{1.852} = 0.040^{1.852}$. $\ln 0.040 = -3.2189$; $\times 1.852 = -5.9614$; $e^{-5.9614} = 0.002576$.

$C^{1.852} = 120^{1.852}$. $\ln 120 = 4.7875$; $\times 1.852 = 8.8665$; $e^{8.8665} = 7102$.

Numerator constant $= 10.67 \times 0.002576 \times 1500 = 41.23$

So $h_f = \dfrac{41.23}{7102 \times D^{4.87}} = \dfrac{0.005806}{D^{4.87}}$

Check D = 250 mm = 0.25 m: $D^{4.87} = 0.25^{4.87}$: $\ln 0.25 = -1.3863$; $\times 4.87 = -6.7513$; $e^{-6.7513} = 0.001168$.

$$h_f = \frac{0.005806}{0.001168} = 4.97\,\text{m}$$

Wait — recompute carefully. $0.005806/0.001168 = 4.97$. This is less than 18 m, so even 250 mm appears adequate. Let us verify the velocity to choose sensibly.

Velocity in 250 mm pipe: $A = \dfrac{\pi}{4}(0.25)^2 = 0.04909\,\text{m}^2$; $v = 0.040/0.04909 = 0.815\,\text{m/s}$ — acceptable (0.6–2.0 m/s range).

Selected diameter: 250 mm, giving head loss $h_f = 4.97\,\text{m}$.

Residual head check at B:

$$H_B = 30.0 - 4.97 = 25.0\,\text{m} \;\ge\; 12\,\text{m} \quad ✓$$

Conclusion: A 250 mm diameter pipe is sufficient. Residual head at B $= \mathbf{25.0\,\text{m}}$ (well above the 12 m requirement), and the velocity of 0.82 m/s lies within the recommended range, so the 250 mm pipe is the economical choice. Larger diameters (300/350 mm) give even lower head loss but are unnecessarily costly and produce low velocities that risk sediment deposition.

Given: $Q = 25\,\text{L/s} = 0.025\,\text{m}^3/\text{s}$, static head $H_s = 45\,\text{m}$, losses $h_L = 6\,\text{m}$, $\eta_{pump} = 0.70$, $\eta_{motor} = 0.90$, $\rho g = 9810\,\text{N/m}^3$.

(a) Heads and powers

Total manometric head:

$$H = H_s + h_L = 45 + 6 = \mathbf{51\,\text{m}}$$

Water (hydraulic) power:

$$P_w = \rho g Q H = 9810 \times 0.025 \times 51 = 12{,}508\,\text{W} = \mathbf{12.51\,\text{kW}}$$

Shaft (brake) power = water power / pump efficiency:

$$P_{shaft} = \frac{12.51}{0.70} = \mathbf{17.87\,\text{kW}}$$

Input electrical power = shaft power / motor efficiency:

$$P_{elec} = \frac{17.87}{0.90} = \mathbf{19.86\,\text{kW}}$$

(Equivalently $P_{elec} = P_w/(\eta_{pump}\eta_{motor}) = 12.51/(0.70\times0.90) = 12.51/0.63 = 19.86\,\text{kW}$.)

(b) Daily energy and cost

Running 8 h/day:

$$E = P_{elec} \times t = 19.86 \times 8 = \mathbf{158.9\,\text{kWh/day}}$$

Daily cost:

$$\text{Cost} = 158.9 \times 11 = \mathbf{NPR\;1{,}748/\text{day}}$$

(Approximately NPR 1,750 per day; ~NPR 52,400 per 30-day month.)

(a) Daily alum requirement

$Q = 10\,\text{MLD} = 10 \times 10^6\,\text{L/day}$, dose $= 28\,\text{mg/L}$.

$$\text{Alum} = \frac{10\times10^6 \times 28}{10^6}\,\text{mg/day in g?}$$

Work in consistent units: mass $= Q \times \text{dose} = 10\times10^6\,\text{L} \times 28\,\text{mg/L} = 2.8\times10^8\,\text{mg} = 280\,\text{kg/day}$.

$$\boxed{= \mathbf{280\,\text{kg/day}}}$$

(b) Solution feed rate (5% solution)

A 5% solution contains 5 kg alum per 100 kg solution. With density ≈ 1000 kg/m³ (1 kg/L), 100 kg solution = 100 L, so it carries 50 g alum per litre (i.e. 0.05 kg/L).

Daily solution volume:

$$V = \frac{280\,\text{kg/day}}{0.05\,\text{kg/L}} = 5{,}600\,\text{L/day}$$

Feed rate:

$$= \frac{5600}{24} = \mathbf{233.3\,\text{L/hour}}$$

(c) Need for alkalinity: Alum [$\text{Al}_2(\text{SO}_4)_3$] hydrolyses to form the insoluble floc $\text{Al(OH)}_3$, releasing $\text{H}^+$ (acidity). Natural alkalinity (bicarbonates) neutralises this acid and supplies $\text{OH}^-$ to complete floc formation. Without enough alkalinity, the pH drops, floc formation is poor, and supplementary alkali (lime/soda ash) must be added.

(a) Bleaching powder and residual chlorine

Volume $V = 500\,\text{m}^3 = 500{,}000\,\text{L}$, applied dose $= 2.5\,\text{mg/L}$.

Mass of chlorine required:

$$\text{Cl}_2 = 500{,}000 \times 2.5 = 1{,}250{,}000\,\text{mg} = 1.25\,\text{kg}$$

Bleaching powder has 30% available chlorine:

$$\text{Bleaching powder} = \frac{1.25}{0.30} = \mathbf{4.17\,\text{kg}}$$

Residual chlorine = applied dose − demand:

$$= 2.5 - 0.6 = \mathbf{1.9\,\text{mg/L}}$$

(This exceeds the typical minimum free residual of 0.2 mg/L at the consumer end, providing a safe margin.)

(b) Factors affecting chlorination efficiency (any two):

1. Contact time — longer contact gives more complete disinfection (CT value).
2. pH — at higher pH the weaker hypochlorite ion ($\text{OCl}^-$) dominates over the more potent hypochlorous acid ($\text{HOCl}$), reducing efficiency.
3. Temperature — higher temperature speeds kill rate.
4. Turbidity/organic matter — shields organisms and exerts chlorine demand.

(a) Sources of water

(b) Water-quality parameters

• Turbidity: A measure of the cloudiness of water caused by suspended/colloidal matter; expressed in NTU. Significance: high turbidity shields pathogens from disinfection, is aesthetically objectionable, and indicates need for coagulation/filtration. Drinking-water guideline ≤ 5 NTU (ideally < 1 NTU).

• Total hardness: The total concentration of multivalent cations, mainly $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$, expressed as mg/L as CaCO₃. Significance: hard water causes scale in pipes/boilers and excess soap consumption; very soft water can be corrosive. Acceptable up to ~500 mg/L.

• Most Probable Number (MPN): A statistical estimate of the number of coliform organisms per 100 mL of water obtained from the multiple-tube fermentation test. Significance: it is the principal bacteriological indicator of faecal contamination; for drinking water the MPN of E. coli should be 0 per 100 mL.

(a) Distribution layouts (any two)

1. Dead-end / Tree system — a main with branching sub-mains and laterals.

   Main ----+----+----+----
            |    |    |
          lateral lateral

Advantage: simple to design and lay; less pipe length, so lower initial cost. (Disadvantage: stagnant dead ends, large area cut off during repair.)

2. Grid-iron / Ring system — interconnected loops with no dead ends.

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

Advantage: water reaches any point from more than one direction, so head loss is low and supply continues during repairs; better for fire-fighting.

(Other valid systems: radial, ring system.)

(b) Service reservoir capacity

Average daily demand $= 3{,}600\,\text{m}^3$.

Maximum-day demand:

$$Q_{max} = 1.5 \times 3600 = 5{,}400\,\text{m}^3$$

Balancing storage at 40% of max-day demand:

$$\text{Capacity} = 0.40 \times 5400 = \mathbf{2{,}160\,\text{m}^3}$$

(To this, fire reserve and breakdown/emergency storage are usually added in practice; here the balancing component alone is 2,160 m³.)

(a) Design population (arithmetic increase)

Annual increase $= 2.0\%$ of present population $= 0.02 \times 2400 = 48$ persons/year.

$$P_{15} = P_0 + (\text{rate} \times n) = 2400 + (48 \times 15) = 2400 + 720 = \mathbf{3{,}120 \text{ persons}}$$

(b) Average daily demand

Demand $= 45\,\text{lpcd}$:

$$Q_{avg} = \frac{3120 \times 45}{1000} = \mathbf{140.4\,\text{m}^3/\text{day}}$$

(c) Peak flow

Average flow in L/s:

$$Q_{avg} = \frac{140.4 \times 1000}{86400} = 1.625\,\text{L/s}$$

Peak flow (peak factor 3.0):

$$Q_{peak} = 3.0 \times 1.625 = \mathbf{4.88\,\text{L/s}}$$

The distribution network and tap-stand sizing must be based on this peak demand of ~4.9 L/s, while the source intake and transmission main are sized on the average (or maximum-day) demand of 140.4 m³/day.

(a) Aeration

Aeration is the process of bringing water into intimate contact with air to exchange gases and add dissolved oxygen. Objectives (any three):

1. To add dissolved oxygen and oxidise dissolved iron and manganese so they can be filtered out.
2. To remove dissolved gases such as CO₂, H₂S, and other taste/odour-causing volatile substances.
3. To drive off volatile organic compounds and improve the palatability (freshness) of the water.

(b) Theoretical oxygen demand for iron removal

Flow $Q = 2\,\text{MLD} = 2\times10^6\,\text{L/day}$; iron $= 6\,\text{mg/L}$.

Daily iron load:

$$\text{Fe} = \frac{2\times10^6 \times 6}{10^6}\,\text{kg/day} = 12\,\text{kg/day}$$

Oxygen demand at 0.14 mg O₂ per mg Fe²⁺:

$$\text{O}_2 = 0.14 \times 12 = \mathbf{1.68\,\text{kg/day}}$$

Thus about 1.68 kg of oxygen per day must be supplied by aeration to oxidise the dissolved iron (in practice a surplus is provided to ensure complete oxidation and to satisfy other oxygen demands).