Nepal Engineering Council · Civil Engineering · Chapter 6
Water Supply, Sanitation and Environment
Tap an option to check it. Wrong picks show the right answer and the hint.
187 questions in 6 syllabus topics.
6.1 Water sources, water quality and water demand
31 questions · ACiE0601
1. Which of the following sources of water generally has the highest turbidity and therefore needs the most treatment?
Show hintHide hint
Think of what runoff carries into streams.
Show answerHide answer
Answer: D. River water during the monsoon
Rivers carry large silt and clay loads in flood, so surface water in the rainy season is the most turbid; spring and deep groundwater are naturally filtered.
2. A hill village in Nepal has a perennial spring located well above the settlement. Which feature makes this source the most economical to develop?
Show hintHide hint
Consider energy cost of lifting water.
Show answerHide answer
Answer: B. Water can be supplied by gravity without pumping
A source at higher elevation than the demand point allows gravity flow, eliminating pumping cost; storage and disinfection are still normally required.
3. Water that rises to the ground surface under hydrostatic pressure from a confined aquifer through a fissure is called
Show hintHide hint
The water is under pressure from confinement.
Show answerHide answer
Answer: B. an artesian spring
A confined aquifer under pressure discharges as an artesian spring; depression springs occur where the ground dips below the water table.
4. An aquifer bounded both above and below by impermeable layers and holding water under pressure is a
Show hintHide hint
The water has no free surface.
Show answerHide answer
Answer: A. confined aquifer
A confined (artesian) aquifer lies between two impermeable strata; an unconfined aquifer has a free water table.
5. An infiltration gallery is best described as
Show hintHide hint
The word 'gallery' implies a horizontal collector.
Show answerHide answer
Answer: C. a horizontal perforated or porous conduit laid below the water table in a permeable stratum
Infiltration galleries collect groundwater (often along a river bank) through perforated pipes or open-jointed conduits laid horizontally below the water table.
6. A catchment of 5 km² receives 1500 mm of annual rainfall. If the runoff coefficient is 0.4, the annual runoff volume available is about
Show hintHide hint
Convert km² to m² and mm to m first.
Show answerHide answer
Answer: D. 3.0 × 10⁶ m³
V = C × P × A = 0.4 × 1.5 m × 5×10⁶ m² = 3.00e+06 m³.
7. The yield of a spring is measured by filling a 20 litre bucket in 8 seconds. If the average demand is 45 litres per capita per day, the population this spring can serve is about
Show hintHide hint
Convert L/s to L/day.
Show answerHide answer
Answer: A. 4800
Q = 20/8 = 2.5 L/s = 216,000 L/day; 216,000/45 = 4800 persons.
8. Colloidal particles in water are those whose size lies approximately between
Show hintHide hint
They are larger than molecules but too small to settle.
Show answerHide answer
Answer: D. 1 nm and 1 µm
Dissolved matter is below about 1 nm, colloids span 1 nm–1 µm, and suspended matter is larger than about 1 µm.
9. Colloidal particles do not settle by gravity mainly because
Show hintHide hint
Think of surface charge.
Show answerHide answer
Answer: B. like electrical charges on their surfaces cause mutual repulsion
Surface charges of the same sign repel each other and the particles stay stable; coagulation neutralises these charges.
10. Turbidity of drinking water is measured with a nephelometer and expressed in
Show hintHide hint
Light scattering instrument.
Show answerHide answer
Answer: A. NTU
A nephelometer measures scattered light; results are given in Nephelometric Turbidity Units. Hazen units are for colour.
11. A 100 mL sample of water is evaporated and the dry residue weighs 75 mg. The total solids concentration is
Show hintHide hint
Convert 100 mL to litres.
Show answerHide answer
Answer: C. 750 mg/L
TS = 75 mg / 0.1 L = 750 mg/L.
12. Temporary (carbonate) hardness of water is due to the presence of
Show hintHide hint
It disappears on boiling.
Show answerHide answer
Answer: A. bicarbonates of calcium and magnesium
Ca and Mg bicarbonates precipitate as carbonates on boiling, so this hardness is called temporary.
13. Permanent (non-carbonate) hardness is caused mainly by
Show hintHide hint
Not removed by boiling.
Show answerHide answer
Answer: B. sulphates and chlorides of calcium and magnesium
Chlorides and sulphates of Ca and Mg do not precipitate on boiling and need lime-soda or ion exchange for removal.
14. A water sample contains 80 mg/L of calcium and 24 mg/L of magnesium. The total hardness expressed as CaCO₃ is about
Show hintHide hint
Convert each ion to CaCO₃ equivalents (50/equivalent weight).
Show answerHide answer
Answer: C. 299 mg/L
H = 2.497×Ca + 4.118×Mg = 2.497×80 + 4.118×24 = 299 mg/L as CaCO₃.
15. In the titration of alkalinity, the phenolphthalein end point and the methyl orange end point correspond to pH values of about
Show hintHide hint
Phenolphthalein turns colourless at the higher pH.
Show answerHide answer
Answer: A. 8.3 and 4.5
Phenolphthalein alkalinity ends at pH ≈ 8.3; total alkalinity (methyl orange) ends at pH ≈ 4.5.
16. 100 mL of a sample requires 12 mL of 0.02 N H₂SO₄ to reach the methyl orange end point. The total alkalinity is
Show hintHide hint
Use 50,000 as the factor.
Show answerHide answer
Answer: C. 120 mg/L as CaCO₃
Alkalinity = (mL × N × 50,000)/sample mL = 12×0.02×50,000/100 = 120 mg/L.
17. A water has total hardness 250 mg/L and total alkalinity 180 mg/L, both as CaCO₃. Its carbonate hardness and non-carbonate hardness are respectively
Show hintHide hint
Carbonate hardness cannot exceed alkalinity.
Show answerHide answer
Answer: D. 180 mg/L and 70 mg/L
Carbonate hardness is the lesser of hardness and alkalinity (180); non-carbonate = 250 − 180 = 70 mg/L.
18. The main domestic nuisance caused by hard water is
Show hintHide hint
Think of washing and boilers.
Show answerHide answer
Answer: A. increased consumption of soap and scale formation in hot-water systems
Ca and Mg react with soap to form scum and deposit scale on heating surfaces.
19. The organism most widely used as an indicator of faecal pollution in drinking water is
Show hintHide hint
Intestinal bacterium.
Show answerHide answer
Answer: D. Escherichia coli
E. coli lives in the intestines of warm-blooded animals, so its presence signals recent faecal contamination.
20. Which of the following is a water-borne disease?
Show hintHide hint
Transmitted by drinking contaminated water.
Show answerHide answer
Answer: D. Cholera
Cholera is transmitted by ingesting contaminated water; malaria is vector-borne, scabies is water-washed and schistosomiasis is water-based.
21. Methaemoglobinaemia (blue-baby syndrome) in infants is associated with excess
Show hintHide hint
Think of fertiliser and sewage contamination.
Show answerHide answer
Answer: B. nitrate in drinking water
Nitrate is reduced to nitrite in an infant's gut and impairs oxygen transport in blood.
22. Amoebic dysentery is caused by
Show hintHide hint
The name 'amoeba' is the clue.
Show answerHide answer
Answer: C. a protozoan, Entamoeba histolytica
Entamoeba histolytica is a protozoan; cholera and typhoid are bacterial and infectious hepatitis is viral.
23. The E. coli count required in drinking water supplied to consumers is
Show hintHide hint
Safe water must be free of faecal indicators.
Show answerHide answer
Answer: C. zero in any 100 mL sample
All drinking water standards require E. coli (thermotolerant coliforms) to be absent in 100 mL.
24. In Nepal, the free residual chlorine commonly specified at the consumer's tap after disinfection is
Show hintHide hint
A very small residual is enough.
Show answerHide answer
Answer: A. 0.1 to 0.2 mg/L
A small residual (about 0.1–0.2 mg/L) is kept to protect against recontamination without causing a chlorine taste.
25. According to IS 1172, which domestic use accounts for the largest share of the per capita demand of about 135 L/day?
Show hintHide hint
It is more than flushing.
Show answerHide answer
Answer: B. Bathing
Of the ~135 lpcd, bathing takes about 55 L, flushing about 30 L, washing of clothes about 20 L; drinking and cooking about 5 L each.
26. A town of 20,000 people has an average demand of 150 L/c/d. Using a peak factor of 2.7, the peak hourly demand expressed in MLD is
Show hintHide hint
Average first, then apply the factor.
Show answerHide answer
Answer: C. 8.1
Average = 20,000×150 = 3.0 ML/d; peak = 2.7×3.0 = 8.1 MLD.
27. Census populations of a town were 20,000 in 1990 and 26,000 in 2000. By the arithmetic increase method the population in 2020 is
Show hintHide hint
Add the same increment each decade.
Show answerHide answer
Answer: A. 38,000
Increase per decade = 6,000; 2020 is two decades after 2000: 26,000 + 2×6,000 = 38,000.
28. A town's population rises from 20,000 to 24,200 in one decade. By the geometric progression method the population after the next decade is about
Show hintHide hint
Use the same ratio again.
Show answerHide answer
Answer: B. 29,282
Growth ratio = 24,200/20,000 = 1.21; next value = 24,200×1.21 = 29,282.
29. Census populations in three successive decades were 10,000, 13,000 and 17,000. By the incremental increase method the population two decades after the last census is
Show hintHide hint
P = P₀ + nx + n(n+1)/2·y.
Show answerHide answer
Answer: B. 27,000
Mean increase x = (3000+4000)/2 = 3500; mean incremental increase y = 1000. P = 17,000 + 2(3500) + [2×3/2](1000) = 27,000.
30. Using Kuichling's formula Q = 3182√P (L/min, P in thousands), the fire demand for a town of population 16,000 is
Show hintHide hint
P must be taken in thousands.
Show answerHide answer
Answer: D. 12,728 L/min
Q = 3182 × √16 = 12,728 L/min.
31. A water utility produces 5,000 m³/day and bills for 3,500 m³/day. The unaccounted-for water is
Show hintHide hint
Compare the loss with production.
Show answerHide answer
Answer: C. 30%
UFW = (5000−3500)/5000 = 30%.
6.2 Intake and distribution systems
31 questions · ACiE0602
32. A river intake is best located on the concave (outer) side of a bend because there
Show hintHide hint
Silt is deposited where velocity is low.
Show answerHide answer
Answer: D. the water is deeper and silt deposition is least
On the outer bank the thalweg lies close to the shore, giving adequate depth throughout the year with little silting, whereas the inner side deposits sediment.
33. With respect to a town on a river, the intake should be located
Show hintHide hint
Think of where the water is cleanest.
Show answerHide answer
Answer: B. upstream of the town and of any sewage or industrial outfall
Locating the intake upstream avoids drawing polluted water into the supply.
34. A Tyrolean (drop) intake, commonly used on steep mountain streams in Nepal, takes water through
Show hintHide hint
The word 'drop' suggests where the inlet is.
Show answerHide answer
Answer: B. a trench with a rack in the bed of the stream
The bottom rack lets water fall into a collecting trench while boulders and large debris pass over, suiting high-gradient, bed-load rich streams.
35. The overflow pipe in a spring-collection chamber (spring box) is provided to
Show hintHide hint
Think about what happens when yield exceeds demand.
Show answerHide answer
Answer: B. discharge surplus water and prevent back-pressure on the spring eye
Excess yield leaves through the overflow so the water level in the box cannot build up and divert or damage the spring.
36. Which pipe material is most suitable for long, flexible, corrosion-free rural supply lines with fusion-welded joints?
Show hintHide hint
Plastic and flexible.
Show answerHide answer
Answer: A. High-density polyethylene (HDPE)
HDPE is flexible, light, corrosion-resistant and available in long coils; joints are made by butt or electro-fusion.
37. Steel pipes are preferred in water supply when
Show hintHide hint
Strength is high, corrosion resistance is not.
Show answerHide answer
Answer: A. high pressures or large diameters are required and a protective lining/coating is provided
Steel has high strength and can be welded, but it corrodes and must be coated and lined.
38. An expansion joint is provided in a pipeline in order to
Show hintHide hint
Think of heat.
Show answerHide answer
Answer: A. allow for longitudinal movement due to temperature changes
It permits axial expansion and contraction, preventing buckling or tension in rigid pipelines.
39. Which of the following valves is NOT suited to continuous throttling of flow?
Show hintHide hint
It is an isolating valve.
Show answerHide answer
Answer: B. Sluice (gate) valve
A gate valve is meant to be fully open or fully closed; partial opening causes vibration and rapid wear of the gate, while globe and needle valves regulate flow.
40. A reflux (non-return) valve in a pipeline is installed to
Show hintHide hint
Name tells you what it refuses to do.
Show answerHide answer
Answer: C. allow flow in one direction only
A swing flap closes under reverse flow and prevents backflow, for example on a pump delivery line.
41. An air valve in a gravity pipeline is generally provided at
Show hintHide hint
Air rises.
Show answerHide answer
Answer: C. the summits (high points) of the line
Air collects at high points and reduces the flow; air valves expel it and also admit air when the line is emptied.
42. A scour (wash-out) valve in a pipeline is provided at
Show hintHide hint
Silt settles at the bottom.
Show answerHide answer
Answer: C. the lowest points (depressions) of the line
Sediment collects in depressions, so the scour valve allows it to be flushed out and the pipe to be emptied.
43. A pipeline has a maximum working head of 60 m. If the field hydrostatic test pressure is taken as 1.5 times the working pressure, the test head is
Show hintHide hint
Multiply by the safety factor.
Show answerHide answer
Answer: A. 90 m
Test head = 1.5 × 60 = 90 m of water (about 0.88 MPa).
44. A pipe at a point 120 m below the water surface of a closed (no-flow) supply tank carries a static pressure of about
Show hintHide hint
p = ρgh with h in metres.
Show answerHide answer
Answer: C. 1.18 MPa
p = ρgh = 1000×9.81×120 = 1.18 MPa (≈ 11.8 bar).
45. A break pressure tank (BPT) is provided in a gravity pipeline in hilly terrain to
Show hintHide hint
Think of excess pressure in steep lines.
Show answerHide answer
Answer: B. reduce the static head to atmospheric pressure so pipes are not over-stressed
The BPT dissipates the head by opening the flow to the atmosphere; the next pipe section starts from zero static pressure.
46. A gravity main descends 250 m in total. If the pipe is rated for a maximum static head of 100 m, the minimum number of break pressure tanks required is
Show hintHide hint
Count sections first, then the tanks between them.
Show answerHide answer
Answer: D. 2
250/100 = 2.5, so 3 pipe sections are needed, i.e. 2 BPTs between them.
47. The functions of a service (distribution) reservoir include
Show hintHide hint
It is a storage unit.
Show answerHide answer
Answer: B. meeting peak demand, maintaining pressure and holding emergency/fire reserve
Service reservoirs equalise the difference between supply and demand, maintain pressure and give an emergency reserve.
48. For a town on undulating ground, the best location for the service reservoir is
Show hintHide hint
Height gives pressure.
Show answerHide answer
Answer: A. a high point near the demand centre
A high location gives gravity pressure to the network and short mains to the demand centre.
49. The capacity of a balancing reservoir is usually determined from
Show hintHide hint
Plot cumulative quantities.
Show answerHide answer
Answer: D. a mass curve of cumulative supply and cumulative demand
The maximum vertical intercepts between the cumulative supply and cumulative demand curves give the storage needed.
50. Water is supplied uniformly over 24 h to a service reservoir. The demand in six consecutive 4-hour periods is 10%, 15%, 30%, 25%, 15% and 5% of the 2000 m³ daily total. The balancing storage needed is about
Show hintHide hint
Add the largest surplus and largest deficit.
Show answerHide answer
Answer: C. 433 m³
Supply per period = 16.67%. Cumulative (supply − demand) = 6.67, 8.33, -5.00, -13.33, -11.67, 0.00%. Storage = max − min = 8.33 + 13.33 = 21.67% of 2000 = 433 m³.
51. A square ground-level service reservoir must hold 450 m³ with a water depth of 3 m. The side of the tank should be about
Show hintHide hint
Capacity = area × depth.
Show answerHide answer
Answer: D. 12.2 m
Area = 450/3 = 150 m²; side = √150 = 12.2 m.
52. A town uses 2400 m³ of water a day. A service reservoir is to hold balancing storage of 25% of the daily demand, a fire reserve of 150 m³ and an emergency reserve of 10% of the daily demand. The total capacity is
Show hintHide hint
Sum the three components.
Show answerHide answer
Answer: A. 990 m³
0.25×2400 = 600; 0.10×2400 = 240; total = 600 + 150 + 240 = 990 m³.
53. The principal disadvantage of the dead-end (tree) distribution system is
Show hintHide hint
No circulation.
Show answerHide answer
Answer: D. stagnation of water at the dead ends and large pressure drop at the far end when a section is closed
Water can come from one direction only, so dead ends gather sediment and stagnant water; closing a pipe cuts off all downstream consumers.
54. In a grid-iron or ring distribution system, compared with a dead-end system,
Show hintHide hint
Loops give alternatives.
Show answerHide answer
Answer: C. water can reach any point by more than one route, so supply is more reliable and stagnation is less
Looped networks supply from more than one direction, improving reliability and pressure, but analysis (e.g. Hardy Cross) is more complex.
55. The Hardy Cross method for analysing a pipe network is based on the conditions that
Show hintHide hint
Continuity plus energy.
Show answerHide answer
Answer: B. the algebraic sum of flows at each junction is zero and the algebraic sum of head losses around each loop is zero
Continuity at nodes and energy balance around loops are used with iterative flow corrections.
56. In a loop, pipe 1 (r = 100) carries 0.05 m³/s clockwise and pipe 2 (r = 200) carries 0.03 m³/s anticlockwise. Taking h = rQ², the first Hardy Cross correction ΔQ (clockwise positive) is
Show hintHide hint
ΔQ = −Σh/Σ(2r|Q|).
Show answerHide answer
Answer: C. −0.0032 m³/s
Σh = 100(0.05)² − 200(0.03)² = 0.07; Σ2rQ = 2(100×0.05 + 200×0.03) = 22; ΔQ = −0.07/22 = −0.0032 m³/s (anticlockwise).
57. Using the Hazen–Williams formula h = 10.67 L Q¹·⁸⁵² / (C¹·⁸⁵² D⁴·⁸⁷), the head loss in 1000 m of 200 mm pipe carrying 30 L/s with C = 120 is about
Show hintHide hint
Use SI units: Q in m³/s, D in m.
Show answerHide answer
Answer: D. 5.8 m
h = 10.67×1000×0.03^1.852 /(120^1.852 × 0.2^4.87) = 5.8 m.
58. The Darcy–Weisbach head loss in 500 m of 150 mm pipe with f = 0.02 and a velocity of 1.2 m/s is about
Show hintHide hint
h = fLV²/2gD.
Show answerHide answer
Answer: A. 4.9 m
h = fLV²/(2gD) = 0.02×500×1.44/(19.62×0.15) = 4.89 m.
59. A branch main must carry 20 L/s at a design velocity of 1.0 m/s. The theoretical pipe diameter is about
Show hintHide hint
A = Q/V then area of a circle.
Show answerHide answer
Answer: D. 160 mm
A = Q/V = 0.02 m²; D = √(4A/π) = 160 mm, so a 160 mm nominal pipe would be chosen.
60. Compared with a single pipe carrying a total flow Q, two identical pipes of the same length laid in parallel carry Q/2 each. Taking head loss proportional to Q², the head loss in each pipe is
Show hintHide hint
Square the flow ratio.
Show answerHide answer
Answer: A. one-quarter of the head loss in the single pipe
h ∝ Q², so halving the flow reduces the head loss to (1/2)² = 1/4.
61. A tank water level is at 150 m, the ground level at a node is 120 m, and the head loss from the tank to the node is 18 m. The residual head at the node is
Show hintHide hint
HGL minus ground level.
Show answerHide answer
Answer: B. 12 m
Residual head = 150 − 18 − 120 = 12 m.
62. A disadvantage of intermittent water supply is that
Show hintHide hint
What happens to an empty leaking pipe?
Show answerHide answer
Answer: C. the pipes empty and may draw in polluted water through leaks when pressure is lost
When the system is depressurised, contaminated groundwater or sewage can enter through joints, and consumers must store water.
6.3 Water treatment process and technologies
31 questions · ACiE0603
63. In a conventional surface-water treatment plant, the usual sequence of units is
Show hintHide hint
Disinfection is the last barrier.
Show answerHide answer
Answer: B. screening → plain sedimentation → coagulation/flocculation → sedimentation → filtration → disinfection
Coarse matter is removed first, then settleable solids, then colloids by coagulation, followed by filtration of residual floc and finally disinfection.
64. Coarse screens (bar racks) at a river intake are provided mainly to
Show hintHide hint
They stop objects, not chemicals.
Show answerHide answer
Answer: C. stop floating debris, branches and large objects from entering the plant
Screens protect pumps and downstream units by retaining large floating and suspended objects.
65. Plain sedimentation (without chemicals) is mainly effective for removing
Show hintHide hint
Chemicals are not added.
Show answerHide answer
Answer: D. heavier silt and sand particles by gravity
Discrete particles of larger size settle by gravity alone; colloids need coagulation first.
66. Using Stokes' law v = g(Sₛ − 1)d²/(18ν), the settling velocity of a sand particle (Sₛ = 2.65, d = 0.05 mm) in water at 20 °C (ν = 1.01 × 10⁻⁶ m²/s) is about
Show hintHide hint
Convert d to metres before squaring.
Show answerHide answer
Answer: D. 2.23 mm/s
v = 9.81×1.65×(5×10⁻⁵)²/(18×1.01×10⁻⁶) = 0.00223 m/s = 2.23 mm/s (Re = 0.11 < 1, so Stokes' law is valid).
67. A rectangular sedimentation tank 20 m long and 6 m wide treats 3600 m³/day. Its surface overflow rate is
Show hintHide hint
Overflow rate = Q/plan area.
Show answerHide answer
Answer: A. 30 m³/m²/day
SOR = Q/A = 3600/(20×6) = 30 m³/m²/day; this equals the settling velocity of the smallest particle fully removed.
68. In an ideal horizontal-flow sedimentation tank, the removal of a given discrete particle depends on
Show hintHide hint
Overflow rate.
Show answerHide answer
Answer: D. the surface area of the tank and the flow rate, not on the depth
Hazen's theory shows removal depends on the overflow rate Q/A; depth affects only the detention time.
69. Alum [Al₂(SO₄)₃·14H₂O] used as coagulant reacts with natural alkalinity. For this reason
Show hintHide hint
The reaction consumes bicarbonate.
Show answerHide answer
Answer: C. water of low alkalinity needs lime or soda ash to be added with alum
Alum consumes bicarbonate alkalinity (about 0.5 mg/L as CaCO₃ per mg/L of alum) and lowers pH; if alkalinity is insufficient, lime or soda ash is added.
70. A raw water is dosed with 40 mg/L of alum. Taking 0.5 mg/L of alkalinity (as CaCO₃) consumed per mg/L of alum, the alkalinity consumed is
Show hintHide hint
Multiply by the stoichiometric ratio.
Show answerHide answer
Answer: B. 20 mg/L
Alkalinity used = 40 × 0.5 = 20 mg/L as CaCO₃.
71. When ferrous sulphate (copperas) is used as a coagulant it is generally applied together with
Show hintHide hint
Needs high pH.
Show answerHide answer
Answer: C. lime, to raise the pH so that ferrous iron is oxidised to ferric hydroxide floc
Copperas needs a high pH (about 8.5 or above) to form an insoluble ferric hydroxide floc.
72. The jar test is performed to
Show hintHide hint
Many jars with different doses.
Show answerHide answer
Answer: B. determine the optimum coagulant dose and pH for a given raw water
A series of jars is dosed at different coagulant doses and stirred to find the dose giving the best floc and clarity.
73. In a jar test, 4 mL of a 10 g/L alum stock solution is added to a 1 litre jar. The alum dose is
Show hintHide hint
1 g/L equals 1 mg/mL.
Show answerHide answer
Answer: B. 40 mg/L
10 g/L = 10 mg/mL; 4 mL × 10 mg/mL = 40 mg in 1 L = 40 mg/L.
74. A plant treats 0.1 m³/s and the optimum alum dose is 30 mg/L. The alum requirement per day is
Show hintHide hint
mg/L equals g/m³.
Show answerHide answer
Answer: A. 259.2 kg
Flow = 8640 m³/d; 30 g/m³ × 8640 = 259,200 g = 259.2 kg/day.
75. The purpose of flocculation is to
Show hintHide hint
Slow mixing, not rapid.
Show answerHide answer
Answer: C. promote gentle collisions of destabilised particles so that large settleable flocs form
After rapid mixing, slow stirring brings the micro-flocs together to form large flocs without breaking them.
76. The velocity gradient is G = √(P/μV). For a flocculator of volume 20 m³ dissipating 18 W in water of μ = 1.002 × 10⁻³ N·s/m², G is about
Show hintHide hint
Take the square root at the end.
Show answerHide answer
Answer: A. 30 s⁻¹
G = √(18/(1.002×10⁻³×20)) = √898 = 30 s⁻¹, within the typical flocculation range of 20–75 s⁻¹.
77. Tube settlers or lamella (plate) settlers improve sedimentation because they
Show hintHide hint
Settling area.
Show answerHide answer
Answer: D. greatly increase the effective settling area and reduce the settling depth
Inclined tubes provide a large settling surface in a small plan area, so higher overflow rates can be used.
78. Slow sand filters are characterised by
Show hintHide hint
Think 'slow' and 'biological'.
Show answerHide answer
Answer: A. a filtration rate of about 0.1–0.4 m/h and a biological layer (schmutzdecke) on the sand surface
SSFs work at low rates; the biologically active schmutzdecke removes bacteria and organic matter. They are cleaned by scraping the top layer.
79. A slow sand filter is required to treat 1.2 ML/day at a filtration rate of 0.2 m/h. The net filter area required is
Show hintHide hint
Area = flow / rate.
Show answerHide answer
Answer: C. 250 m²
Q = 1200 m³/d = 50 m³/h; A = 50/0.2 = 250 m².
80. A rapid sand filter plant must supply 5000 m³/day of filtered water. 4% of the filtered water is used for backwashing, the filter works 24 h/day and the filtration rate is 5 m/h. The total filter area required is
Show hintHide hint
Allow for wash water in the gross flow.
Show answerHide answer
Answer: C. 43.3 m²
Gross flow = 5000×1.04 = 5200 m³/d = 216.7 m³/h; A = 216.7/5 = 43.3 m².
81. A rapid sand filter of area 20 m² is backwashed at 36 m/h for 10 minutes. The volume of wash water used is
Show hintHide hint
Convert minutes to hours.
Show answerHide answer
Answer: B. 120 m³
Q = 36 m/h × 20 m² = 720 m³/h; for 10 min = 720×10/60 = 120 m³.
82. The effective size of filter sand is the
Show hintHide hint
Subscript denotes percent finer.
Show answerHide answer
Answer: D. sieve size through which 10% of the sand by weight passes (D₁₀)
Effective size is D₁₀; the uniformity coefficient is D₆₀/D₁₀.
83. If the sieve analysis of filter sand gives D₆₀ = 0.84 mm and D₁₀ = 0.42 mm, the uniformity coefficient is
Show hintHide hint
Divide the larger by the smaller.
Show answerHide answer
Answer: B. 2.0
UC = D₆₀/D₁₀ = 0.84/0.42 = 2.0.
84. The dose of chlorine applied to a water is 2.5 mg/L and the residual after the contact period is 0.5 mg/L. The chlorine demand is
Show hintHide hint
Demand is what the water consumes.
Show answerHide answer
Answer: A. 2.0 mg/L
Demand = dose − residual = 2.5 − 0.5 = 2.0 mg/L.
85. In break-point chlorination, the break point is the point at which
Show hintHide hint
After chloramines are destroyed.
Show answerHide answer
Answer: B. the chlorine residual falls to a minimum after the destruction of chloramines and further dosing gives free residual
Beyond the break point, additional chlorine appears as free available chlorine.
86. 500 m³ of water must be dosed with 2 mg/L of chlorine using bleaching powder containing 25% available chlorine. The quantity of bleaching powder required is
Show hintHide hint
Divide by the available fraction.
Show answerHide answer
Answer: A. 4 kg
Chlorine = 500 m³ × 2 g/m³ = 1 kg; bleaching powder = 1/0.25 = 4 kg.
87. Disinfection by ozone or ultraviolet radiation differs from chlorination in that they
Show hintHide hint
Think about protection after treatment.
Show answerHide answer
Answer: D. leave no residual disinfectant in the distribution system
Ozone decomposes rapidly and UV acts only at the point of application, so a chlorine residual may still be needed.
88. A water contains 150 mg/L of carbonate hardness, all due to calcium (as CaCO₃). Based on Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ + 2H₂O, the theoretical hydrated lime Ca(OH)₂ needed is about
Show hintHide hint
Molecular weights 74 and 100.
Show answerHide answer
Answer: A. 111 mg/L
Ca(OH)₂ (74) : CaCO₃ equivalent (100) so 150 × 0.74 = 111 mg/L.
89. Non-carbonate hardness of 80 mg/L as CaCO₃ is removed by the lime-soda process using soda ash (Na₂CO₃, MW 106). The soda ash required is about
Show hintHide hint
Ratio of 106 to 100.
Show answerHide answer
Answer: C. 85 mg/L
Na₂CO₃ (106) per CaCO₃ equivalent (100): 80 × 1.06 = 84.8 mg/L.
90. A zeolite (ion exchange) softener, when exhausted, is regenerated with
Show hintHide hint
Common salt.
Show answerHide answer
Answer: D. a strong sodium chloride (brine) solution
Brine replaces the Ca²⁺ and Mg²⁺ held on the resin with Na⁺, restoring its capacity.
91. Aeration of raw water is carried out to
Show hintHide hint
Gas exchange.
Show answerHide answer
Answer: A. remove dissolved gases like CO₂ and H₂S and to oxidise dissolved iron
Aeration strips volatile gases, adds oxygen and oxidises Fe²⁺ and Mn²⁺ so they can be removed by sedimentation/filtration.
92. Dissolved iron in groundwater is commonly removed by
Show hintHide hint
Oxidise first, then separate.
Show answerHide answer
Answer: B. aeration followed by sedimentation and filtration
Aeration oxidises soluble ferrous iron to insoluble ferric hydroxide which is then removed by settling and filtration.
93. The Kanchan arsenic filter, widely promoted in the Terai region of Nepal, removes arsenic mainly by
Show hintHide hint
Iron is involved.
Show answerHide answer
Answer: C. adsorption on iron hydroxide formed from rusting iron nails above a sand layer
Iron nails rust and the resulting iron hydroxide adsorbs arsenic, which is retained in the filter.
6.4 Design and construction of sewers
31 questions · ACiE0604
94. In the design of a sewerage system, the quantity of dry-weather sewage is usually taken as about
Show hintHide hint
Not all supplied water returns.
Show answerHide answer
Answer: B. 80% of the per capita water supply
Some water is lost to lawn watering, evaporation and leakage, so about 75–80% of the supplied water reaches the sewers.
95. A town of 15,000 people has a water supply of 150 L/c/d. Taking 80% of it as sewage, the average dry-weather flow is
Show hintHide hint
Multiply population, per capita demand and 0.8.
Show answerHide answer
Answer: A. 1.8 ML/d
15,000 × 150 × 0.8 = 1,800,000 L/d = 1.8 ML/d.
96. Harmon's peak factor is M = 1 + 14/(4 + √P) where P is population in thousands. For a population of 36,000 the peak factor is
Show hintHide hint
√36 = 6.
Show answerHide answer
Answer: B. 2.4
M = 1 + 14/(4 + √36) = 1 + 14/10 = 2.4.
97. Infiltration of groundwater into a sewer is greater when
Show hintHide hint
Needs a way in and pressure.
Show answerHide answer
Answer: C. the water table is high and the sewer joints are poor
Groundwater leaks in through defective joints and cracks below the water table, increasing the flow.
98. The rational formula Q = CIA/360 gives Q in m³/s when I is in mm/h and A in hectares. For C = 0.7, I = 60 mm/h and A = 12 ha, the storm runoff is
Show hintHide hint
Check the units of the formula.
Show answerHide answer
Answer: C. 1.4 m³/s
Q = 0.7×60×12/360 = 1.4 m³/s.
99. An area of 10 ha consists of 4 ha of paved surface (C = 0.9) and 6 ha of lawns (C = 0.3). The composite runoff coefficient is
Show hintHide hint
Weight by area.
Show answerHide answer
Answer: A. 0.54
C = (4×0.9 + 6×0.3)/10 = 0.54.
100. In a separate system of sewerage,
Show hintHide hint
The name gives it away.
Show answerHide answer
Answer: C. sanitary sewage and storm water are carried in different sewers
A separate system has one sewer for domestic/industrial wastewater and another (often open drains) for storm water.
101. An advantage of a separate system over a combined system is that
Show hintHide hint
Treatment load.
Show answerHide answer
Answer: D. only the sanitary sewage needs to be treated, so the treatment works are smaller
Storm water can be discharged to natural drains without treatment, so the volumes sent to the treatment works are smaller.
102. A disadvantage of a combined sewer system is that
Show hintHide hint
Think of heavy rain.
Show answerHide answer
Answer: D. during heavy storms the sewers and treatment works are overloaded, requiring storm overflows
The large storm flow is far greater than dry-weather flow, so overflow structures are needed, sending diluted sewage to the stream.
103. The dry-weather flow (DWF) in a sewer includes
Show hintHide hint
No rain involved.
Show answerHide answer
Answer: C. domestic sewage, industrial wastewater and groundwater infiltration
DWF is the flow in the absence of rainfall: domestic and trade wastes plus infiltration.
104. The minimum velocity in a sewer is prescribed to ____ and the maximum velocity is limited to ____.
Show hintHide hint
Both extremes have a problem.
Show answerHide answer
Answer: C. prevent deposition of solids (self-cleansing); prevent abrasion/scouring of the sewer invert
Too low a velocity allows solids to settle and produce septic conditions; too high a velocity abrades the sewer with grit.
105. A commonly adopted maximum (non-scouring) velocity for concrete sewers is about
Show hintHide hint
A few metres per second.
Show answerHide answer
Answer: A. 3 m/s
Velocities above about 2.4–3 m/s abrade concrete and clay invert surfaces.
106. A circular sewer of diameter 300 mm laid at a slope of 1 in 200 flows full. Using Manning's n = 0.013, the discharge is about
Show hintHide hint
R for a full circular pipe is D/4.
Show answerHide answer
Answer: D. 68 L/s
R = D/4 = 0.075 m; V = (1/0.013)(0.075)^(2/3)(0.005)^½ = 0.97 m/s; Q = V×πD²/4 = 68 L/s.
107. A 200 mm circular sewer (n = 0.013, running full) must achieve a self-cleansing velocity of 0.6 m/s. The minimum slope required is about
Show hintHide hint
Rearrange Manning for S.
Show answerHide answer
Answer: A. 1 in 303
S = (Vn/R^(2/3))² = (0.6×0.013/0.05^(2/3))² = 0.00330 ≈ 1 in 303.
108. For a circular sewer, the velocity when flowing half full is
Show hintHide hint
Compare hydraulic radii.
Show answerHide answer
Answer: A. equal to the velocity when flowing full
Hydraulic radius is the same (D/4) at half depth and full depth, so with the same slope the velocity is the same; discharge is half.
109. In a circular sewer, the discharge is a maximum at a depth of flow of about
Show hintHide hint
Slightly below full.
Show answerHide answer
Answer: B. 0.95 of the diameter
Q is maximum at d/D ≈ 0.95 (about 7% more than full-flow discharge); velocity is maximum at ≈ 0.81 D.
110. Sewers are not designed to flow completely full mainly in order to
Show hintHide hint
Allow some air space.
Show answerHide answer
Answer: D. provide ventilation and a margin for flow variations above the design flow
Partial flow leaves an air space for ventilation and extra capacity for peaks and infiltration.
111. An egg-shaped sewer is generally preferred for combined systems because it
Show hintHide hint
Low flows must still be self-cleansing.
Show answerHide answer
Answer: B. maintains a reasonable depth and velocity at low (dry-weather) flows
The narrow bottom gives adequate velocity during dry-weather flow while the wide upper part carries storm flow.
112. Which sewer material is most resistant to acidic wastewater and to corrosion by sewer gases?
Show hintHide hint
Ceramic.
Show answerHide answer
Answer: C. Vitrified clay (stoneware)
Glazed vitrified clay pipes resist chemical attack and abrasion, though they are brittle and heavy.
113. Crown corrosion in concrete sewers is caused by
Show hintHide hint
Sulphur gas.
Show answerHide answer
Answer: C. hydrogen sulphide being oxidised to sulphuric acid on the moist pipe crown
Anaerobic slime produces H₂S which is oxidised by bacteria to H₂SO₄ on the exposed pipe wall, attacking the concrete.
114. While laying a sewer pipeline in a trench, the pipes are laid
Show hintHide hint
Against the direction of flow.
Show answerHide answer
Answer: D. starting from the downstream end with the socket facing upstream
Laying from the downstream end with sockets facing upstream lets the spigot of each new pipe enter the previous socket and any water drains away.
115. The gradient of a sewer during laying is controlled using
Show hintHide hint
Rails and rods.
Show answerHide answer
Answer: A. sight rails and boning rods (travellers)
Sight rails are fixed above the trench; a boning rod of calculated length is used to check the invert level of each pipe.
116. Which test is used to check the water-tightness of a newly laid gravity sewer?
Show hintHide hint
Fill it with water.
Show answerHide answer
Answer: B. Hydrostatic (water) test
The section between two manholes is filled with water to a specified head and the leakage measured; air tests are also used.
117. Manholes in a sewer line are provided at
Show hintHide hint
Access points where something changes.
Show answerHide answer
Answer: D. changes in alignment, gradient and diameter, at junctions and at suitable intervals
They give access for inspection, cleaning and rodding, and are located wherever sewers change direction, slope, size or meet.
118. A drop manhole is provided when
Show hintHide hint
Level difference.
Show answerHide answer
Answer: A. a branch sewer enters a manhole at a level much higher than the main sewer
A vertical drop pipe lets the branch discharge to the main invert without erosion or splashing, usually when the difference exceeds about 0.6 m.
119. Catch basins (street inlets) in a storm sewer system serve to
Show hintHide hint
Sump collects solids.
Show answerHide answer
Answer: B. admit surface runoff and retain grit and heavy debris
Runoff enters through grating, and a sump retains silt to prevent clogging of the storm sewer.
120. An inverted siphon in a sewer system is provided to
Show hintHide hint
Goes under an obstacle.
Show answerHide answer
Answer: D. carry the sewer under a river, railway or depression where the gradient line cannot be maintained
A depressed sewer runs under pressure below the hydraulic gradient across obstacles and then rises again.
121. A trap fitted at the connection of a house drain with a sewer prevents
Show hintHide hint
A water seal.
Show answerHide answer
Answer: C. sewer gases from entering the house
A water seal in the bend blocks passage of foul gas into the building.
122. Ground level at an upstream manhole is 100.00 m and the sewer invert is 2.0 m below it. The sewer is 90 m long at a slope of 1 in 100. If the ground level at the downstream manhole is 99.20 m, the depth of the invert below ground there is
Show hintHide hint
Fall = length × slope.
Show answerHide answer
Answer: B. 2.10 m
Downstream invert = 98.00 − 0.90 = 97.10 m; depth = 99.20 − 97.10 = 2.10 m.
123. A combined sewer carries a dry-weather flow of 0.05 m³/s plus a storm flow of 1.40 m³/s. An overflow chamber passes only 3 times the DWF to the treatment plant. The flow diverted to the overflow is
Show hintHide hint
Subtract flow sent to treatment.
Show answerHide answer
Answer: A. 1.30 m³/s
Total = 1.45 m³/s; to treatment = 3×0.05 = 0.15; overflow = 1.45 − 0.15 = 1.30 m³/s.
124. A 300 mm circular sewer laid at 1 in 200 (n = 0.013) has a full-flow discharge of about 68 L/s. When it flows half full, its discharge is about
Show hintHide hint
Same velocity, half the area.
Show answerHide answer
Answer: B. 34 L/s
At half depth the area is half and the velocity equals the full-flow velocity, so Q = 0.5 × 68 = 34 L/s.
6.5 Treatment and disposal of wastewater
31 questions · ACiE0605
125. The correct order of nitrogen forms as organic matter in sewage is progressively oxidised is
Show hintHide hint
Nitrification sequence.
Show answerHide answer
Answer: D. organic nitrogen → ammonia → nitrite → nitrate
Hydrolysis gives ammonia, which is oxidised to nitrite and then nitrate in nitrification; nitrate indicates a well-stabilised effluent.
126. The standard BOD test is carried out by incubating the sample for
Show hintHide hint
Name 'BOD₅' gives the days.
Show answerHide answer
Answer: B. 5 days at 20 °C
Standard BOD₅ is measured after 5 days at 20 °C in the dark.
127. 10 mL of sewage is diluted to 300 mL with aerated dilution water. The initial dissolved oxygen is 8.6 mg/L and after 5 days at 20 °C it is 4.4 mg/L. The 5-day BOD of the sewage is
Show hintHide hint
Multiply the oxygen drop by the dilution factor.
Show answerHide answer
Answer: C. 126 mg/L
BOD₅ = (DOᵢ − DOf) × dilution factor = (8.6 − 4.4) × 300/10 = 126 mg/L.
128. The 5-day BOD of a wastewater is 200 mg/L and the BOD rate constant is k = 0.23 d⁻¹ (base e). The ultimate BOD is about
Show hintHide hint
BOD_t = L₀(1 − e^(−kt)).
Show answerHide answer
Answer: D. 293 mg/L
L₀ = BOD₅/(1 − e^(−5k)) = 200/(1 − e^(−1.15)) = 200/0.683 = 293 mg/L.
129. Chemical oxygen demand (COD) of a wastewater is normally
Show hintHide hint
The chemical oxidant is stronger than bacteria.
Show answerHide answer
Answer: D. greater than or equal to its BOD, since it also oxidises non-biodegradable organics
COD uses a strong chemical oxidant (potassium dichromate) that oxidises both biodegradable and many non-biodegradable compounds, so COD ≥ BOD.
130. The main purpose of a grit chamber in a sewage treatment plant is to
Show hintHide hint
Inorganic solids.
Show answerHide answer
Answer: B. remove sand, gravel and other heavy inorganic particles
Grit would abrade pumps and settle in channels and digesters; grit chambers remove it without settling the organic matter.
131. In a horizontal-flow grit chamber, the flow velocity is controlled at about 0.3 m/s so that
Show hintHide hint
Selective settling.
Show answerHide answer
Answer: B. grit settles while the lighter organic matter is carried forward
At about 0.3 m/s organic particles stay in suspension but the heavier inorganic grit settles.
132. A horizontal-flow grit chamber handles 0.6 m³/s at a flow-through velocity of 0.3 m/s with a depth of 1.0 m. If the grit has to settle at 0.025 m/s, the required length is
Show hintHide hint
Time to settle one depth × horizontal velocity.
Show answerHide answer
Answer: C. 12 m
Settling time = depth/vs = 1.0/0.025 = 40 s; L = v × t = 0.3 × 40 = 12 m (width = Q/(v·d) = 2 m).
133. Primary sedimentation of raw domestic sewage typically removes about
Show hintHide hint
About half to two-thirds of solids.
Show answerHide answer
Answer: A. 60% of suspended solids and 30–35% of BOD
Plain sedimentation removes settleable solids and associated BOD but not dissolved or colloidal organics.
134. In a trickling filter, organic matter in the sewage is removed mainly by
Show hintHide hint
Biofilm.
Show answerHide answer
Answer: A. a biological film (zoogleal slime) growing on the filter media
Sewage trickles over a bed of stones or plastic media and microorganisms in the slime layer oxidise the organic matter.
135. A trickling filter of diameter 20 m treats 2000 m³/day of settled sewage. The hydraulic loading is about
Show hintHide hint
Use plan area of the circular bed.
Show answerHide answer
Answer: C. 6.4 m³/m²/day
Area = π×20²/4 = 314.2 m²; loading = 2000/314.2 = 6.37 m³/m²/day.
136. In the activated sludge process, part of the settled sludge is returned to the aeration tank in order to
Show hintHide hint
The organisms must be retained.
Show answerHide answer
Answer: A. maintain a high concentration of active microorganisms in the aeration tank
Return activated sludge seeds the incoming sewage with microorganisms, maintaining the required MLSS.
137. A 30-minute settling test on mixed liquor with MLSS 2500 mg/L gives a settled sludge volume of 250 mL/L. The sludge volume index is
Show hintHide hint
SVI = V₃₀/MLSS in mL per gram.
Show answerHide answer
Answer: B. 100 mL/g
SVI = (settled volume mL/L × 1000)/MLSS mg/L = 250×1000/2500 = 100 mL/g.
138. An aeration tank of volume 1250 m³ receives 5000 m³/day of settled sewage with BOD 200 mg/L. The MLSS is 2000 mg/L. The food-to-microorganism ratio is
Show hintHide hint
Mass of BOD per day / mass of MLSS.
Show answerHide answer
Answer: A. 0.4 per day
F/M = Q·S₀/(V·X) = 5000×200/(1250×2000) = 0.4 d⁻¹.
139. For the same aeration tank (volume 1250 m³) and flow 5000 m³/day, the hydraulic retention time is
Show hintHide hint
Volume over flow.
Show answerHide answer
Answer: A. 6.0 hours
HRT = V/Q = 1250/5000 = 0.25 d = 6.0 h.
140. A very high sludge volume index (for example above 200 mL/g) in an activated sludge plant indicates
Show hintHide hint
Large volume per gram.
Show answerHide answer
Answer: A. poorly settling, bulking sludge
High SVI means the sludge occupies a large volume, usually due to filamentous organisms, and it settles poorly in the clarifier.
141. In a facultative oxidation (stabilisation) pond, the oxygen needed by bacteria in the upper layer is supplied mainly by
Show hintHide hint
Sunlight is involved.
Show answerHide answer
Answer: C. photosynthesis by algae and surface re-aeration
Algae produce oxygen in sunlight and the bacteria use it to oxidise organic matter, releasing CO₂ for the algae.
142. A treatment plant reduces BOD₅ from 250 mg/L to 25 mg/L. The BOD removal efficiency is
Show hintHide hint
Removed over initial.
Show answerHide answer
Answer: A. 90%
Efficiency = (250 − 25)/250 × 100 = 90%.
143. The dissolved oxygen deficit in a stream is defined as
Show hintHide hint
Saturation is the reference.
Show answerHide answer
Answer: B. saturation DO minus actual DO
D = DOsat − DO. In the Streeter–Phelps equation the deficit is the quantity that is traced downstream.
144. In the oxygen sag curve downstream of a sewage outfall, the critical point is the point at which
Show hintHide hint
Lowest point of the sag.
Show answerHide answer
Answer: D. the dissolved oxygen is a minimum (the deficit is a maximum)
At the critical point the rate of deoxygenation equals the rate of re-aeration and the DO is lowest.
145. In the Streeter–Phelps analysis, k_d = 0.2 d⁻¹, k_r = 0.4 d⁻¹ (base e), initial ultimate BOD L_a = 20 mg/L and initial deficit D_a = 0. The critical time t_c = ln(k_r/k_d)/(k_r − k_d) is about
Show hintHide hint
ln 2 = 0.693.
Show answerHide answer
Answer: D. 3.47 days
t_c = ln(2)/0.2 = 3.47 days.
146. For the same stream (k_d = 0.2 d⁻¹, k_r = 0.4 d⁻¹, L_a = 20 mg/L, D_a = 0), the critical deficit D_c = (k_d/k_r)·L_a·e^(−k_d t_c) is
Show hintHide hint
e^(−0.693) = 0.5.
Show answerHide answer
Answer: B. 5.0 mg/L
With t_c = 3.47 d, e^(−0.2×3.47) = 0.5, so D_c = 0.5 × 20 × 0.5 = 5.0 mg/L.
147. Disposal of effluent by dilution into a river is acceptable only when
Show hintHide hint
Assimilative capacity.
Show answerHide answer
Answer: C. the river has adequate flow and DO to absorb the BOD load without falling below standards
The stream's assimilative capacity depends on dilution ratio, DO, temperature and velocity.
148. Land treatment of sewage by irrigation (sewage farming) can lead to 'sewage sickness' of the soil which is
Show hintHide hint
Soil pores blocked.
Show answerHide answer
Answer: B. clogging of soil pores by organic matter, making the soil unable to absorb more sewage
Overloading with sewage clogs soil voids with grease and solids, stopping aeration and infiltration.
149. The gas produced by anaerobic digestion of sewage sludge consists mainly of
Show hintHide hint
Combustible gas.
Show answerHide answer
Answer: D. methane (about 60–65%) and carbon dioxide
Methane-forming bacteria produce a combustible biogas, often used to heat the digester or generate power.
150. A sludge contains 1000 kg of dry solids per day at 4% solids concentration (specific gravity 1.02). The daily volume of sludge is about
Show hintHide hint
Mass of solids / (density × solids fraction).
Show answerHide answer
Answer: C. 24.5 m³
V = M/(ρ·s) = 1000/(1020 × 0.04) = 24.5 m³/day.
151. A sludge of 10 m³ with 98% moisture is thickened until its moisture content is 95%. The resulting volume is
Show hintHide hint
Solids content stays the same.
Show answerHide answer
Answer: B. 4 m³
Dry solids are conserved: 10×(100−98) = V×(100−95), so V = 4 m³.
152. Of the usual solid waste disposal methods, the greatest volume reduction (about 90%) is achieved by
Show hintHide hint
Heat.
Show answerHide answer
Answer: D. incineration
Burning converts waste to ash and flue gas, reducing the volume by about 90%.
153. A septic tank is to serve 20 users at 120 L/person/day with a 24-hour liquid detention. Sludge storage is 30 L/person/year with desludging every 2 years. The required capacity is
Show hintHide hint
Add liquid volume and sludge volume.
Show answerHide answer
Answer: A. 3.6 m³
Liquid = 20×120 = 2400 L = 2.4 m³; sludge = 20×30×2 = 1200 L = 1.2 m³; total = 3.6 m³.
154. Effluent from a septic tank is usually disposed of by
Show hintHide hint
Still polluted.
Show answerHide answer
Answer: C. a soak pit or dispersion trench, since it is still highly polluted
Septic tank effluent retains a significant BOD and pathogens, so it is percolated through a soak pit or leach field.
155. A pour-flush latrine differs from a ventilated improved pit (VIP) latrine in that
Show hintHide hint
Water seal vs vent.
Show answerHide answer
Answer: C. a water seal prevents odours and flies, whereas the VIP uses a vent pipe with a fly screen
Pour-flush toilets have a water-seal trap; VIP toilets control smell and flies through a screened vent pipe.
6.6 Concept of environmental assessment
32 questions · ACiE0606
156. Sustainable development is best defined as development that
Show hintHide hint
Think of future generations.
Show answerHide answer
Answer: D. meets the needs of the present without compromising the ability of future generations to meet their needs
This is the Brundtland Commission (1987) definition.
157. The primary purpose of Environmental Impact Assessment (EIA) is to
Show hintHide hint
It comes before the decision, not after.
Show answerHide answer
Answer: A. identify and predict the environmental effects of a proposed project before decisions are taken, and suggest mitigation
EIA is a planning tool that predicts significant impacts in advance so that they can be avoided, reduced or compensated.
158. Which sequence correctly lists the main steps of the EIA process?
Show hintHide hint
Decide whether, then what, then collect data.
Show answerHide answer
Answer: D. Screening → scoping → baseline study → impact prediction and evaluation → mitigation and EMP → monitoring
Screening decides whether EIA is needed; scoping defines the issues; baseline data, prediction, mitigation/EMP and monitoring follow.
159. In the EIA process, 'screening' is the step that
Show hintHide hint
First step: does it need assessment?
Show answerHide answer
Answer: B. decides whether a proposed project requires BES, IEE or EIA
Screening classifies the proposal against the legal schedules to decide the level of assessment needed.
160. In an EIA, 'scoping' refers to the step in which
Show hintHide hint
It focuses the study.
Show answerHide answer
Answer: C. the significant issues and impacts to be studied are identified and the terms of reference are prepared
Scoping narrows the study to the key issues and results in the terms of reference (ToR).
161. An Environmental Management Plan (EMP) in an EIA report is mainly a plan that
Show hintHide hint
Management of impacts.
Show answerHide answer
Answer: C. lists the mitigation measures, monitoring programme, responsibilities and budget
The EMP turns the predicted impacts into specific actions, monitoring and institutional responsibilities.
162. As compared with an EIA, an Initial Environmental Examination (IEE) is
Show hintHide hint
The word 'initial'.
Show answerHide answer
Answer: A. a less detailed study for smaller projects with less significant impacts
IEE is a screening-level, shorter study; EIA is a detailed study for projects with potentially significant adverse impacts.
163. In Nepal's Environment Protection Act, 2019 (2076 BS), the Brief Environmental Study (BES) applies to
Show hintHide hint
The smallest tier.
Show answerHide answer
Answer: B. small proposals with limited, easily managed environmental impacts
The Act provides a three-tier system: BES for small-scale proposals, IEE for medium and EIA for projects with potentially significant impacts.
164. The Environment Protection Act in force in Nepal at present, which replaced the Environment Protection Act of 1997 (2053 BS), was enacted in
Show hintHide hint
Replaced the 1997 Act in the late 2010s.
Show answerHide answer
Answer: A. 2019 (2076 BS)
The Environment Protection Act 2076 (2019) replaced the Environment Protection Act 2053 (1997); the Environment Protection Rules 2077 (2020) followed.
165. The baseline environmental data collected for an EIA in Nepal usually cover
Show hintHide hint
Four environmental components.
Show answerHide answer
Answer: A. physical, biological, socio-economic and cultural environments
Baseline surveys characterise the existing physical, biological, socio-economic and cultural conditions of the project area.
166. Which of the following is the correct hierarchy of environmental mitigation?
Show hintHide hint
Prevention is better than cure.
Show answerHide answer
Answer: A. Avoid → minimise → rehabilitate/restore → compensate
The preferred order is to avoid the impact first, then reduce it, restore damaged areas and finally compensate for residual impacts.
167. The Leopold matrix is an EIA method that
Show hintHide hint
Matrix = rows and columns.
Show answerHide answer
Answer: D. lists project actions along one axis and environmental characteristics along the other to indicate interactions
It is a two-dimensional matrix with 100 actions and 88 environmental characteristics, in which each interaction is rated for magnitude and importance.
168. The original Leopold matrix has 100 project actions and 88 environmental characteristics. The number of possible interaction cells is
Show hintHide hint
Multiply rows by columns.
Show answerHide answer
Answer: C. 8,800
100 × 88 = 8,800 cells.
169. The 'network' method of impact assessment is particularly useful for showing
Show hintHide hint
Chain of effects.
Show answerHide answer
Answer: C. secondary and higher-order (indirect) impacts through cause–effect chains
Networks trace how a primary impact triggers a chain of secondary and tertiary effects.
170. In the Battelle environmental evaluation system, a parameter of importance weight 30 PIU falls in environmental quality from 0.8 to 0.6 after the project. The change in environmental impact units is
Show hintHide hint
Weight × change in quality.
Show answerHide answer
Answer: B. -6 EIU
EIU = PIU × EQ; change = 30×(0.6 − 0.8) = -6 EIU (a loss).
171. On a road slope in the hills of Nepal, low-cost control of surface erosion is usually obtained by
Show hintHide hint
Use living plants.
Show answerHide answer
Answer: D. bio-engineering, using planted grasses, shrubs and trees with drainage
Vegetation binds the soil, reduces rain impact and runoff, and complements drainage and small structures.
172. A strategic environmental assessment (SEA) differs from EIA in that it is applied to
Show hintHide hint
The word 'strategic'.
Show answerHide answer
Answer: C. policies, plans and programmes rather than individual projects
SEA addresses environmental implications at the policy, plan and programme level, before individual projects are defined.
173. Which one of the following is a man-made (anthropogenic) disaster?
Show hintHide hint
Caused by human error or activity.
Show answerHide answer
Answer: D. Industrial chemical leakage
Industrial accidents, fires and stampedes are human-caused; earthquakes, rainfall-induced landslides and GLOF are natural hazards.
174. The disaster management cycle consists of
Show hintHide hint
Before, during and after the event.
Show answerHide answer
Answer: A. mitigation, preparedness, response and recovery
The four phases are mitigation, preparedness, response and recovery (rehabilitation and reconstruction).
175. In disaster risk assessment, risk is commonly understood as a function of
Show hintHide hint
Hazard is only one part.
Show answerHide answer
Answer: C. hazard, exposure and vulnerability
Risk increases with the hazard, the elements exposed and their vulnerability, and decreases with coping capacity.
176. Which of the following is a non-structural flood mitigation measure?
Show hintHide hint
No construction involved.
Show answerHide answer
Answer: B. Flood-plain zoning and flood forecasting/early warning
Zoning and warning systems reduce damage without altering the flood itself, whereas embankments, dams and channel works are structural.
177. The most effective and generally the first measure for stabilising a landslide-prone slope is
Show hintHide hint
Water is the main trigger.
Show answerHide answer
Answer: D. controlling surface and subsurface drainage
Water raises pore pressure and weight and reduces shear strength; draining the slope is usually the most economical and effective measure.
178. A glacial lake outburst flood (GLOF) occurs when
Show hintHide hint
Think of ice and moraine dams.
Show answerHide answer
Answer: D. a glacial lake held back by a moraine dam suddenly releases its water
Moraine-dammed lakes can fail by overtopping or piping, producing a sudden destructive flood downstream; artificial lowering and early warning are used for mitigation.
179. The difference between earthquake magnitude and intensity is that
Show hintHide hint
Source vs site.
Show answerHide answer
Answer: B. magnitude measures the energy released at the source, intensity measures the effects felt at a place
Magnitude (Richter/moment) is a single value for the event; Modified Mercalli intensity varies with location.
180. The Gorkha earthquake of 25 April 2015 in Nepal had a moment magnitude of about
Show hintHide hint
Greater than 7.
Show answerHide answer
Answer: A. 7.8
The Mw 7.8 Gorkha earthquake caused about 9,000 deaths and widespread destruction.
181. Each unit increase in earthquake magnitude corresponds to about 31.6 times more energy released (10¹·⁵). A magnitude 7.8 earthquake releases how many times the energy of a magnitude 5.8 earthquake?
Show hintHide hint
Difference is 2 magnitude units.
Show answerHide answer
Answer: C. 1,000
Energy ratio = 10^(1.5×2) = 10³ = 1,000.
182. A 50-year flood is the design flood. The probability that it will be equalled or exceeded at least once in the next 25 years is about
Show hintHide hint
1 − probability of no occurrence.
Show answerHide answer
Answer: C. 40%
P = 1 − (1 − 1/T)ⁿ = 1 − (0.98)²⁵ = 0.397 ≈ 40%.
183. A rainwater harvesting system has a roof of 100 m², annual rainfall 1200 mm and a runoff coefficient of 0.8. The annual volume that can be collected is
Show hintHide hint
Convert mm to metres.
Show answerHide answer
Answer: B. 96 m³
V = A × R × C = 100 m² × 1.2 m × 0.8 = 96 m³.
184. Two identical machines each produce 80 dB at a point. The combined noise level at that point is about
Show hintHide hint
Add intensities, not decibels.
Show answerHide answer
Answer: D. 83 dB
L = 10 log₁₀(2 × 10⁸) = 80 + 10 log 2 = 83 dB.
185. A point noise source produces 80 dB at 10 m. Assuming free-field spherical spreading (−6 dB per doubling of distance), the level at 40 m is about
Show hintHide hint
Two doublings of distance.
Show answerHide answer
Answer: B. 68 dB
ΔL = 20 log(40/10) = 12 dB; level = 80 − 12 = 68 dB.
186. The resisting force along the potential failure surface of a slope is 450 kN/m and the driving force is 360 kN/m. The factor of safety against sliding is
Show hintHide hint
Ratio of resisting to driving force.
Show answerHide answer
Answer: B. 1.25
FS = resisting/driving = 450/360 = 1.25.
187. In Nepal, the Disaster Risk Reduction and Management Act enacted in 2017 (2074 BS) replaced
Show hintHide hint
The older Act dealt with relief.
Show answerHide answer
Answer: A. the Natural Calamity (Relief) Act, 1982
The DRRM Act 2074 established a framework including the National Disaster Risk Reduction and Management Authority, replacing the 1982 relief-focused Act.