Nepal Engineering Council · Civil Engineering · Chapter 8
Hydropower
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184 questions in 6 syllabus topics.
8.1 Planning of hydropower projects
31 questions · ACiE0801
1. The gross (theoretical) hydropower potential of a country is estimated by considering
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Think of the upper bound that ignores every practical limitation.
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Answer: A. all runoff over the full fall of the rivers, assuming 100% efficiency
Gross theoretical potential is computed from the total runoff and the total available fall of all rivers with no losses or constraints, so it is the largest of the three potential estimates.
2. The technical hydropower potential is the part of the gross potential that
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It is a technology filter, not a money filter.
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Answer: B. can be developed with available technology, allowing for site and efficiency constraints
Technical potential removes sites that cannot be built with present technology and applies realistic losses and efficiencies; the economic test is applied afterwards.
3. Economically feasible hydropower potential is the portion of the technical potential for which
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Look for the criterion based on costs versus benefits.
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Answer: C. benefits exceed costs under prevailing economic conditions (BCR ≥ 1)
A site is economically feasible when its discounted benefits at least equal its discounted costs, i.e. BCR ≥ 1 or EIRR above the discount rate.
4. For any country, the correct descending order of hydropower potential estimates is
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Each estimate filters the previous one.
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Answer: D. gross > technical > economic
Each estimate is a subset of the previous one: the economic potential is part of the technical potential, which is part of the gross potential.
5. The theoretical (gross) hydropower potential of Nepal is commonly quoted as about
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The figure is in the tens of thousands of MW.
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Answer: A. 83,000 MW
Nepal's theoretical potential is about 83,000 MW, of which about 42,000 MW is considered economically feasible.
6. The economically feasible hydropower potential of Nepal is commonly stated as about
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It is about half of the theoretical value.
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Answer: C. 42,000 MW
Out of about 83,000 MW of theoretical potential, roughly 42,000 MW is regarded as economically feasible.
7. Among Nepal's major river basins, the one with the largest theoretical hydropower potential is the
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It is the westernmost large river with no major project built on its main stem.
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Answer: D. Karnali basin
The Karnali river system has the largest theoretical potential of the major basins, followed by the Koshi and Narayani basins.
8. The first hydropower plant of Nepal, a 500 kW plant, was commissioned in 1911 at
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It lies on the southern edge of Kathmandu Valley.
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Answer: B. Pharping
Pharping hydropower station (500 kW, 1911) was Nepal's first. Sundarijal (1936), Panauti (1965) and Trishuli (1967) came later.
9. The usual correct order of stages in the development of a hydropower project is
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Each stage needs more data and spends more money than the one before.
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Answer: D. reconnaissance, pre-feasibility, feasibility, detailed design, construction
Studies proceed from coarse to fine: reconnaissance identifies sites, pre-feasibility screens them, feasibility confirms viability and capacity, and detailed design produces drawings and tender documents before construction.
10. The main purpose of a reconnaissance study is to
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It is the cheapest and quickest stage.
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Answer: B. identify and rank potential sites using maps, existing data and brief site visits
Reconnaissance is a desk study with a site visit that screens candidate sites cheaply before any detailed investigation.
11. The optimum installed capacity, detailed hydrological analysis and economic and financial viability of a project are established mainly in the
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It follows pre-feasibility and precedes detailed design.
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Answer: C. feasibility study
The feasibility study fixes the project layout, installed capacity and cost estimate and tests economic and financial viability. Detailed design then refines it.
12. Preparation of final design drawings, specifications and tender documents belongs to the
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It is the last study before contractors are invited.
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Answer: C. detailed engineering design stage
After feasibility confirms the project, detailed design produces the construction drawings, specifications and tender documents.
13. A river basin master plan for hydropower is mainly used to
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The word master suggests the whole basin, not one site.
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Answer: B. identify, compare and rank projects of the whole basin for staged development
A master plan examines all potential sites of a basin together so that development is sequenced rationally and conflicts between uses are avoided.
14. The Water Resources Act of Nepal (1992) is important for hydropower because it
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It concerns who owns water and the right to use it.
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Answer: A. vests the ownership of water resources in the State and requires a licence to use them
The Water Resources Act 1992 declares that ownership of water resources lies with the State and requires a licence for their utilisation. Tariff and regulation are dealt with under other laws.
15. In Nepal, licences for the survey and generation of electricity are issued under the
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Its name says exactly what it governs.
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Answer: D. Electricity Act, 1992
The Electricity Act, 1992 and its Regulation (1993) provide for survey, generation, transmission and distribution licences, handled by the Department of Electricity Development.
16. The Hydropower Development Policy of 1992 is notable mainly because it
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It marked a shift in who could build plants.
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Answer: D. opened hydropower generation to the private sector in Nepal
The 1992 policy, along with the Electricity Act and Water Resources Act of that year, invited private participation in generation. It was later revised in 2001.
17. The Nepal Electricity Authority (NEA) was established in
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It is in the mid-1980s.
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Answer: B. 1985
NEA was formed in 1985 by merging the Department of Electricity, the Nepal Electricity Corporation and related development boards.
18. The Electricity Regulatory Commission of Nepal is mainly responsible for
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It is a regulator, not a builder or generator.
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Answer: D. regulating the electricity sector, including tariff determination
The Commission, established under the Electricity Regulatory Commission Act (2017), regulates the sector, including tariff, and settles disputes.
19. In a Build-Own-Operate-Transfer (BOOT) hydropower model, the developer
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The last word of the abbreviation is the key.
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Answer: C. builds, owns and operates the plant for a licence period and then transfers it to the government
Under BOOT the developer finances and builds the plant, operates it to recover the investment and then hands it over free of cost at the end of the licence period.
20. Nepal's Hydropower Development Policy requires developers to release, as environmental flow, at least
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It is a small percentage of the lowest monthly average.
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Answer: A. 10% of the minimum monthly average flow of the river
The policy requires a minimum release of 10% of the minimum monthly average discharge downstream of the diversion to sustain aquatic life.
21. Hydropower provides roughly what share of the world's electricity generation?
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Roughly one sixth.
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Answer: A. about 16%
Hydropower supplies about one sixth of global electricity and is the largest renewable source.
22. The Three Gorges Dam, the world's largest hydropower station by capacity (about 22,500 MW), is in
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It is on the Yangtze river.
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Answer: B. China
Three Gorges on the Yangtze River in China is the largest power plant of any type by installed capacity. China also leads in total installed hydropower capacity.
23. In Nepal, a hydropower plant of capacity below 100 kW is classified as a
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It is the smallest class.
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Answer: A. micro hydropower plant
In Nepalese practice, plants up to 100 kW are micro hydro, those from 100 kW to 1 MW are mini hydro.
24. The gross theoretical power of a river reach carrying a flow of 50 m³/s over a fall of 100 m (100% efficiency, ρg = 9.81 kN/m³) is about
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Use P = γ Q H with no efficiency.
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Answer: D. 49.05 MW
P = ρ g Q H = 9.81 × 50 × 100 = 49,050 kW ≈ 49.05 MW.
25. A river has a mean flow of 300 m³/s and a usable fall of 50 m. The technical power with an overall efficiency of 85% is about
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Multiply by efficiency as well.
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Answer: A. 125 MW
P = 0.85 × 9.81 × 300 × 50 = 125,077 kW ≈ 125 MW.
26. A 10 MW hydropower plant works at an annual plant factor of 60%. Its yearly energy output is about
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A year has 8760 hours.
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Answer: B. 52,560 MWh
Energy = capacity × 8760 h × plant factor = 10 × 8760 × 0.6 = 52,560 MWh.
27. The present worth of benefits of a project is NPR 1,200 million and the present worth of costs is NPR 1,000 million. The benefit-cost ratio and decision are
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BCR is benefit divided by cost.
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Answer: B. 1.2; economically feasible
BCR = 1200/1000 = 1.2 > 1, so the project is economically feasible. Option inversions give 0.83.
28. A 30 MW hydropower project costs NPR 6,000 million. The cost per installed kW is
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Convert MW to kW first.
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Answer: C. NPR 200,000
Cost per kW = 6,000 × 10⁶ / (30 × 10³ kW) = NPR 200,000/kW.
29. Taking Nepal's theoretical potential as 83,000 MW and economically feasible potential as 42,000 MW, the economic potential is about what percentage of the theoretical potential?
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Divide the smaller figure by the larger.
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Answer: C. 51%
42,000/83,000 = 0.506 ≈ 51%.
30. A plant sells 120 GWh of energy a year at NPR 8.40 per kWh. Its annual energy revenue is
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1 GWh = 10⁶ kWh.
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Answer: C. NPR 1,008 million
Revenue = 120 × 10⁶ kWh × 8.40 = NPR 1,008 × 10⁶ = NPR 1,008 million.
31. A project costing NPR 4,500 million earns a net annual benefit of NPR 600 million. Its simple payback period is
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Divide cost by yearly net benefit.
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Answer: A. 7.5 years
Simple payback = cost / annual net benefit = 4,500/600 = 7.5 years.
8.2 Power and energy potential study
30 questions · ACiE0802
32. A plant has a net head of 85 m, a design discharge of 24 m³/s and an overall efficiency of 88%. The power output is about
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P = η γ Q H.
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Answer: B. 17.6 MW
P = 0.88 × 9.81 × 24 × 85 = 17,611 kW ≈ 17.6 MW (without efficiency it would be 20.0 MW).
33. A flow duration curve is a plot of
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The x-axis is a percentage of time.
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Answer: A. flow against the percentage of time the flow is equalled or exceeded
The flow duration curve ranks flows from highest to lowest and plots each against the percentage of time it is equalled or exceeded.
34. The design discharge of a run-of-river plant in Nepal is commonly taken as Q40. This is the flow that is
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The subscript is a percentage of time.
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Answer: C. equalled or exceeded for about 146 days of the year
Q40 is the flow equalled or exceeded 40% of the time, i.e. 0.40 × 365 ≈ 146 days.
35. Firm (primary) power of a hydropower plant is the power
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Think of reliability.
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Answer: C. that is available practically all the time, based on the minimum dependable flow
Firm power corresponds to the minimum dependable flow (with high dependability) and can be supplied continuously. Surplus power at high flows is secondary power.
36. Secondary power of a run-of-river plant is the power
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It is a bonus beyond the guaranteed part.
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Answer: A. available in excess of firm power during high river flows, without guarantee
Secondary (non-firm) power is the extra energy available when river flow exceeds the firm flow, mostly in the monsoon, and cannot be guaranteed.
37. The average load on a power station is 40 MW and its peak load is 80 MW. The load factor is
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Average divided by peak.
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Answer: C. 0.50
Load factor = average load / peak load = 40/80 = 0.50.
38. A 50 MW plant generates 175,200 MWh in a year. Its annual plant (capacity) factor is
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Compare with the energy at full capacity for 8760 h.
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Answer: B. 0.40
Plant factor = actual energy / (capacity × 8760) = 175,200/(50 × 8760) = 0.40.
39. The ratio of the actual energy produced to the energy that could have been produced if the plant ran at installed capacity throughout the year is called the
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It is defined with respect to installed capacity.
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Answer: C. plant (capacity) factor
Plant factor (capacity factor) compares actual annual energy with that at full installed capacity for 8760 h. Load factor compares average with peak load.
40. Which statement about fixing the installed capacity of a storage plant is most appropriate?
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Think of marginal analysis.
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Answer: A. It is raised until the incremental benefit of the last unit equals its incremental cost
The economic (optimum) capacity is where the additional benefit from an extra kW equals the extra cost; beyond this point additional capacity does not pay.
41. A hydropower plant is classified as low head, medium head or high head mainly on the basis of
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The classification is about head, as the name says.
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Answer: D. the net head available at the plant
Head classification (low, medium, high) is based on the net operating head, which also governs the choice of turbine.
42. A plant working under a net head of 400 m is classed as a
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Compare 400 m with typical class limits of tens of metres.
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Answer: D. high head plant
Heads above about 50 m are generally treated as high head, and 400 m is high under any common classification; Pelton turbines are typically used.
43. A hydropower plant with no significant storage that uses the river flow as it comes is called a
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The name describes the flow regime.
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Answer: D. run-of-river plant
A run-of-river plant has little or no storage, so generation follows the natural river flow.
44. A run-of-river plant provided with a small daily pondage to meet peak demand is called a
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The plant stores for a day only.
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Answer: C. peaking run-of-river (pondage) plant
Daily pondage stores flow during off-peak hours and releases it during peak hours, allowing peaking operation without seasonal storage.
45. In a pumped storage plant, water is
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It acts like a large battery.
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Answer: C. pumped to the upper reservoir during off-peak hours and used for generation at peak hours
Surplus off-peak energy lifts water to the upper reservoir; at peak demand the water is released through the turbines.
46. In a pumped storage scheme the pumping efficiency is 85% and the generating efficiency is 88%. The cycle (round-trip) efficiency is about
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Multiply the two efficiencies.
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Answer: B. 74.8%
Round-trip efficiency = 0.85 × 0.88 = 0.748 ≈ 74.8%.
47. Which of the following is NOT a component of a typical run-of-river hydropower scheme?
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The scheme has no major storage.
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Answer: C. A large storage reservoir with a dead storage zone
Run-of-river schemes divert the river through a weir, intake, settling basin, headrace, forebay, penstock and powerhouse; they have no large storage reservoir.
48. In reservoir terminology, dead storage is the volume
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Water here cannot be released through the power intake.
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Answer: D. below the minimum drawdown level, reserved mainly for sediment accumulation
Dead storage lies below the minimum drawdown level (MDDL) and cannot be drawn for generation; it holds the sediment deposited during the life of the reservoir.
49. The live (useful) storage of a power reservoir lies between
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It is the part that can actually be drawn down.
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Answer: D. full reservoir level (FRL) and minimum drawdown level (MDDL)
Live storage is the volume between FRL and MDDL, available for regulation and power generation.
50. A reservoir has an average surface area of 5 km² between its FRL and MDDL, which are 20 m apart. The live storage is about
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Area times depth, in consistent units.
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Answer: B. 100 million m³
Live storage ≈ mean area × height = 5 × 10⁶ m² × 20 m = 100 × 10⁶ m³.
51. The Rippl mass curve method is used to determine
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It uses cumulative inflow against time.
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Answer: B. the storage required to supply a given demand from varying inflows
The mass curve plots cumulative inflow; the largest departure of the cumulative demand line from the mass curve gives the required storage.
52. Monthly inflows to a reservoir for eight consecutive months are 10, 8, 6, 4, 3, 5, 9 and 12 million m³, and the constant monthly demand is 7 million m³. Assuming the reservoir is full at the start, the minimum required storage is
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Add the successive deficits while the inflow is below demand.
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Answer: D. 10 million m³
Deficits (demand minus inflow) accumulate: Months 3-6 give 1 + 3 + 4 + 2 = 10 million m³ of cumulative deficit; the later surplus months refill the reservoir. Required storage = 10 million m³.
53. A reservoir of 120 million m³ capacity receives 0.60 million m³ of sediment per year and traps 80% of it. The time to fill the reservoir with sediment is
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Only the trapped part fills the reservoir.
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Answer: A. 250 years
Trapped sediment = 0.8 × 0.6 = 0.48 million m³/yr; life = 120/0.48 = 250 years.
54. A storage plant receives a regulated flow averaging 10 m³/s and generates only for 6 hours a day at the peak. With a net head of 200 m and an efficiency of 90%, the peaking capacity is about
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Concentrate the daily volume into 6 hours.
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Answer: B. 70.6 MW
Peak discharge = 10 × 24/6 = 40 m³/s; P = 0.9 × 9.81 × 40 × 200 = 70,632 kW ≈ 70.6 MW.
55. For the plant above (inflow 10 m³/s, peak release 40 m³/s for 6 hours), the daily pondage needed is about
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Only the excess over inflow comes from storage.
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Answer: A. 648,000 m³
Volume = (40 − 10) × 6 × 3600 = 648,000 m³ must be stored from the off-peak inflow.
56. Seasonal storage in a reservoir is intended to
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It works across seasons, not hours.
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Answer: B. store the monsoon surplus for use in the dry season
A seasonal storage reservoir holds high-flow water for release in low-flow months, raising the firm power of the plant.
57. A plant has a gross head of 245 m. Head losses in the tunnel, surge system and penstock total 9.5 m. The net head available at the turbine is
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Subtract losses once from gross head.
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Answer: B. 235.5 m
Net head = gross head − head losses = 245 − 9.5 = 235.5 m.
58. A 30 MW plant operates at an annual plant factor of 55%. Its annual energy generation is about
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Capacity × hours × factor.
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Answer: A. 144.5 GWh
E = 30 MW × 8760 h × 0.55 = 144,540 MWh = 144.5 GWh.
59. Which type of hydropower plant is best suited to meet the peak portion of the daily load curve?
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Peaking needs the ability to store water.
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Answer: D. A storage or pondage plant with large installed capacity operating a few hours daily
Regulated storage or pondage allows water to be held back and released at high discharge during peak hours, while run-of-river plants without pondage serve the base load.
60. A river has a mean annual flow of 18 m³/s. The mean annual runoff volume is about
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Seconds in a year are about 31.5 million.
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Answer: A. 568 million m³
V = 18 × 365 × 86,400 s = 567.6 × 10⁶ m³ ≈ 568 million m³.
61. A reservoir rule curve is used to
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It is an operating guide against the calendar.
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Answer: A. specify the water levels to be kept in the reservoir at different times of the year for operation
A rule curve guides the operator on the target storage at each time of the year, balancing power, flood control and other uses.
8.3 Headworks of storage plants
30 questions · ACiE0803
62. Which of the following is the main feature of a concrete gravity dam?
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Think of how the name describes the stabilising force.
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Answer: A. It resists the water thrust mainly by its own weight
A gravity dam is a massive structure whose self-weight produces the stabilising moment and friction against overturning and sliding. Arch dams carry load to the abutments.
63. An arch dam is most suitable for
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The load is carried sideways.
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Answer: D. a narrow gorge with strong rock abutments
An arch dam transfers thrust to the abutments, so it needs a narrow valley and sound rock on both sides.
64. For a dam site with deep soft alluvium and plentiful local earth and gravel, the most suitable dam type is generally
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Which dam spreads the load widely and uses local soil?
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Answer: C. an earthfill (embankment) dam
Embankment dams impose low bearing pressure and use local materials, so they suit weaker foundations; concrete dams need sound rock.
65. According to ICOLD, a dam is classified as a 'large dam' if its height above the lowest foundation is more than
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It is a modest height of a few storeys.
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Answer: D. 15 m
ICOLD defines a large dam as one with a height of 15 m or more from the lowest general foundation (or 5-15 m with a reservoir over 3 million m³).
66. Which of the following forces is NOT normally considered in the stability analysis of a gravity dam?
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Look for the odd one that is not a dam-specific load.
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Answer: B. Centrifugal force due to river curvature
Usual loads are self-weight, hydrostatic pressure, uplift, silt, earthquake, wave and ice pressure. Centrifugal force from river curvature is not a dam load.
67. Per metre length, the horizontal hydrostatic force on a vertical upstream face of a dam holding 60 m of water is
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Area of the pressure triangle.
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Answer: C. 17,658 kN
P = ½ γ H² = 0.5 × 9.81 × 60² = 17,658 kN per metre, acting at H/3 = 20 m above the base.
68. A gravity dam has a base width of 45 m and 60 m of water at the heel and zero at the toe. If uplift varies linearly from full head at the heel to zero at the toe (no drains), the uplift force per metre is about
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Triangular uplift diagram times the base width.
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Answer: D. 13,244 kN
U = ½ × γ H × B = 0.5 × 9.81 × 60 × 45 = 13,244 kN/m.
69. For a gravity dam, ΣV = 40,000 kN, ΣH = 22,000 kN per metre and the coefficient of friction is 0.70. The factor of safety against sliding (friction only) is
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Friction resistance divided by driving force.
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Answer: D. 1.27
FS = μ ΣV / ΣH = 0.70 × 40,000/22,000 = 1.27.
70. The shear-friction factor (SFF) of a gravity dam differs from the sliding factor of safety because SFF also includes
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The extra term is the shear resistance.
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Answer: A. the shear strength of the joint or foundation along the sliding plane
SFF = (μ ΣV + b τ)/ΣH, where τ is the shear strength and b the area of the shear plane; it accounts for cohesion or shear strength along the plane.
71. The factor of safety of a gravity dam against overturning, taken as the ratio of resisting moment to overturning moment, should normally be at least about
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It is a little more than unity.
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Answer: A. 1.5
Overturning is checked for FS ≥ 1.5 (resultant also kept within the middle third for normal loading).
72. For a dam section 48 m wide at the base, ΣV = 50,000 kN/m and the net moment of all forces about the toe is 1.3 × 10⁶ kN·m/m. The eccentricity of the resultant from the base centre is
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Find where the resultant cuts the base first.
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Answer: B. 2 m
Distance of the resultant from the toe = 1.3 × 10⁶ / 50,000 = 26 m, so e = |26 − 24| = 2 m, which is less than B/6 = 8 m.
73. A dam of base width 40 m has ΣV = 48,000 kN/m and the resultant acts with eccentricity 4 m. The maximum base pressure is
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Use the combined axial and bending formula.
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Answer: B. 1,920 kPa
σmax = (ΣV/B)(1 + 6e/B) = 1200 × (1 + 0.6) = 1,920 kPa; σmin = 480 kPa.
74. The base width of an elementary gravity dam profile (no uplift, no tension) of height 80 m and specific gravity of concrete 2.4 is B = H/√Sc, which is about
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Take the square root of the specific gravity.
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Answer: C. 51.6 m
B = 80/√2.4 = 80/1.549 = 51.6 m.
75. In an earth dam, the chief cause of failure by seepage is
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Seepage can erode the soil internally.
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Answer: A. piping, in which fines are carried out along concentrated seepage paths
Progressive erosion of soil particles by concentrated seepage (piping) forms pipes through the embankment or foundation and may lead to breach. Filters and drains control it.
76. Overtopping failure of an embankment dam is mainly prevented by
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Control the water level before it reaches the crest.
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Answer: B. providing a spillway with adequate capacity and sufficient freeboard
Earth dams cannot withstand flow over the crest; adequate spillway capacity for the design flood and freeboard are the main defences.
77. The function of a chimney drain and toe drain in an earth dam is to
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They must carry seepage water safely out.
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Answer: B. intercept seepage and keep the phreatic line within the dam body
Internal drains collect seepage safely and lower the phreatic surface, preventing piping and slope instability on the downstream side.
78. The condition most critical for the stability of the upstream slope of an earth dam is generally
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The water that supported the slope is removed quickly.
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Answer: B. sudden (rapid) drawdown of the reservoir
During rapid drawdown, pore water pressure in the upstream shell cannot dissipate quickly and the stabilising water load is removed, so the upstream slope is most critical.
79. Lane's weighted creep ratio is (Lv/3 + Lh)/H. For a vertical creep length of 12 m, horizontal creep length of 30 m and net head of 10 m, it is
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Divide vertical lengths by 3 before adding.
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Answer: B. 3.4
(12/3 + 30)/10 = 34/10 = 3.4.
80. Curtain grouting under a dam is mainly done to
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It is placed near the upstream side as a barrier.
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Answer: D. reduce seepage and uplift by forming a low-permeability barrier
A grout curtain along the upstream heel cuts seepage and uplift; consolidation grouting, in contrast, strengthens the shallow foundation rock.
81. Consolidation grouting in the foundation of a concrete dam is carried out to
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It treats the near-surface rock mass.
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Answer: C. improve the strength and reduce the deformability of shallow fractured rock
Consolidation grouting is shallow, closely spaced grouting of fractured rock to improve its strength and stiffness.
82. Drainage holes drilled from the inspection gallery of a concrete gravity dam serve to
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They relieve water pressure under the dam.
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Answer: A. reduce the uplift pressure acting on the base
Drains downstream of the grout curtain relieve water pressure in the foundation, lowering the uplift and increasing stability.
83. In Gordon's formula S = C V √D for the minimum submergence of an intake (C = 0.7245), the velocity at the intake opening is 3 m/s and its height is 3 m. The required submergence is about
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Use the square root of the opening height.
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Answer: C. 3.76 m
S = 0.7245 × 3 × √3 = 0.7245 × 3 × 1.732 = 3.76 m.
84. A trash rack must pass 18 m³/s at an approach velocity of 0.9 m/s through the net opening. If the bars block 20% of the gross area, the gross rack area needed is
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Net area is gross × (1 − blockage).
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Answer: B. 25.0 m²
Net area = 18/0.9 = 20 m²; gross area = 20/0.8 = 25 m².
85. The discharge of an ogee spillway is Q = C L H^1.5. For C = 2.2, effective length 40 m and head 3.5 m, the discharge is about
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Raise the head to the power 3/2.
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Answer: A. 576 m³/s
Q = 2.2 × 40 × 3.5^1.5 = 88 × 6.548 = 576 m³/s.
86. A morning-glory (shaft) spillway is generally preferred for
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It uses a vertical shaft and a tunnel.
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Answer: A. narrow gorge sites where there is no room for a chute or side channel
A shaft spillway has a circular crest at the top of a vertical shaft connected to a tunnel, so it fits narrow valleys with limited space.
87. Water enters a stilling basin at a depth of 0.8 m with velocity 14 m/s. The sequent depth of the hydraulic jump is about
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Compute the Froude number first.
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Answer: C. 5.3 m
Fr = 14/√(9.81 × 0.8) = 5.00; y2 = (0.8/2)(√(1 + 8 × 25.0) − 1) = 5.27 m.
88. A hydraulic jump has conjugate depths of 0.5 m and 3.0 m. The energy loss in the jump is about
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ΔE = (y2 − y1)³ / (4 y1 y2).
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Answer: C. 2.60 m
ΔE = (y2 − y1)³/(4 y1 y2) = 2.5³/(4 × 0.5 × 3.0) = 15.625/6 = 2.60 m.
89. A ski-jump (flip) bucket is used at a spillway to
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The jet is thrown clear into the air.
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Answer: A. throw the jet clear of the dam toe into a plunge pool
A ski-jump bucket throws the high-velocity jet into the air, so that energy is dissipated by air entrainment and in the downstream plunge pool, away from the dam toe. It suits rock foundations.
90. A radial (Tainter) gate is commonly installed on the crest of a spillway because
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The pressure passes through the pivot.
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Answer: D. it needs a small hoisting force as the water thrust passes through the trunnion
In a radial gate the hydrostatic thrust acts through the trunnion pins, so lifting loads are small and no gate slots are required in the piers.
91. An emergency gate placed at a power intake is meant to
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It is a standby, rarely operated gate.
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Answer: D. close the intake in the case of a failure of the service gate or turbine regulation system
Emergency gates are kept fully open and closed only in an emergency, or for inspection when the service gate fails.
8.4 Headworks of run-of-river (ROR) plants
30 questions · ACiE0804
92. Which sequence lists the main headworks components of a typical run-of-river hydropower plant in the direction of flow?
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Coarse sediment is removed before fine sediment.
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Answer: C. weir, intake, gravel trap, settling basin, headrace
The weir raises and diverts the river into the intake; coarse bed load is removed in the gravel trap and fine suspended sediment in the settling basin before the water enters the headrace.
93. To keep bed load out of a side intake, the intake should be located on
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Transverse (helical) currents carry bed material to one bank.
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Answer: C. the outer (concave) bank of a river bend
Spiral flow in a bend carries bed load towards the inner bank; the outer bank has less bed load and deeper water, so the intake is placed there.
94. The sill of a river intake is kept well above the river bed (or the undersluice floor) mainly to
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Bed load moves near the bed.
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Answer: D. prevent coarse bed load from entering the intake
A raised sill, typically 1-1.5 m above the bed, ensures only upper-layer clear water enters while bed load passes through the undersluice.
95. A submerged orifice intake of area 2.5 m² works under an effective head of 0.9 m. With Cd = 0.6, the discharge is about
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Use the orifice equation with the square root of 2gh.
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Answer: D. 6.30 m³/s
Q = Cd A √(2gh) = 0.6 × 2.5 × √(2 × 9.81 × 0.9) = 1.5 × 4.202 = 6.30 m³/s.
96. An undersluice (scouring sluice) adjacent to the intake is provided to
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It scours sediment away from the intake mouth.
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Answer: A. flush bed load and keep the approach to the intake clear
A divide wall separates the undersluice from the intake so that sediment deposited in front of the intake can be flushed downstream.
97. The main function of a gravel trap in a run-of-river scheme is to
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It handles the coarse fraction.
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Answer: C. settle and remove coarse sediment (gravel and cobbles) before it reaches the settling basin
A gravel trap is a short basin or pit just after the intake where velocity is reduced so bed load can be flushed out through a gate.
98. A Tyrolean (drop) intake with a bottom rack is suitable for
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The water drops through a screen while bed load passes over.
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Answer: B. steep, boulder-laden mountain streams
In a Tyrolean intake, water falls through a bar screen in the weir crest, while boulders and gravel pass over it; it suits steep Himalayan streams.
99. The sediment carried in the water body of the flow without touching the bed for most of the time is called
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It is carried in the flow, not on the bed.
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Answer: B. suspended load
Suspended load is carried by turbulence in the flow; bed load rolls, slides or saltates along the bed.
100. The design particle size of a settling basin at a hydropower plant is chosen smaller when
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Velocity increases with head.
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Answer: A. the net head of the plant is higher
Turbine abrasion grows steeply with the jet or flow velocity, which increases with head; high-head plants therefore need finer particles to be settled out.
101. Using Stokes' law w = g(ρs − ρ)d²/(18μ), the settling velocity of a 0.1 mm quartz particle (ρs = 2650 kg/m³) in water (ρ = 1000 kg/m³, μ = 1.0 × 10⁻³ Pa·s) is about
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Remember d is in metres and is squared.
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Answer: D. 9.0 mm/s
w = 9.81 × 1650 × (10⁻⁴)² / (18 × 10⁻³) = 8.99 × 10⁻³ m/s ≈ 9.0 mm/s.
102. Camp's formula gives the critical (non-scouring) velocity in a settling basin as V = a√d, where V is in cm/s and d in mm. For d = 0.2 mm and a = 51, V is about
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Take the square root of the diameter in mm.
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Answer: D. 22.8 cm/s
V = 51 × √0.2 = 51 × 0.447 = 22.8 cm/s. The flow velocity in the basin should be below this to avoid re-suspension.
103. By the Hazen ideal-basin concept, a basin must handle 12 m³/s and remove particles with settling velocity 0.027 m/s. The minimum plan area is
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Surface loading rate equals the settling velocity.
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Answer: C. 444 m²
In an ideal basin the surface loading Q/A equals the settling velocity, so A = Q/w = 12/0.027 = 444 m².
104. A settling basin is 3 m deep. Design particles settle at 0.03 m/s and the flow velocity is 0.25 m/s. The minimum length of the basin (ideal conditions) is
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Length is velocity times settling time.
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Answer: B. 25 m
Settling time t = depth/w = 3/0.03 = 100 s; length = V × t = 0.25 × 100 = 25 m.
105. The actual length of a settling basin is made greater than the ideal length because of
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Real flow is not uniform and quiescent.
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Answer: A. turbulence and short-circuiting
Turbulence and uneven flow distribution reduce the settling efficiency; designers adopt a turbulence or safety factor on the ideal length.
106. In the 'practice approach' of settling basin design, the basin is proportioned to
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It is based on a particle size.
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Answer: A. completely remove all particles larger than a selected design size
The practice approach selects a limiting particle size (depending on head and turbine) and sizes the basin so that particles above it are trapped.
107. In the 'concentration approach' of settling basin design, the basin is designed so that
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It is based on how much sediment may pass, not only on size.
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Answer: A. the sediment concentration in the outflow is within a permissible limit for the turbines
The concentration approach uses the inflow sediment concentration, grain size distribution and the allowed outlet concentration to determine the required trap efficiency and basin size.
108. A river carries 3,000 ppm of sediment into a settling basin of trap efficiency 90%. The sediment concentration in the water leaving the basin is
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Only the untrapped fraction leaves.
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Answer: D. 300 ppm
Outflow concentration = 3,000 × (1 − 0.90) = 300 ppm.
109. Vetter's equation gives the trap efficiency η = 1 − exp(−w A/Q). For w = 0.025 m/s, a basin 40 m long and 8 m wide and Q = 6 m³/s, the efficiency is about
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A is the plan area, B × L.
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Answer: B. 73.6%
wA/Q = 0.025 × 320/6 = 1.333; η = 1 − e^(−1.333) = 1 − 0.264 = 0.736 = 73.6%.
110. A basin receives 10 m³/s of water carrying 2,000 mg/L of sediment and traps 90% of it. If the deposited sediment has a bulk density of 1,500 kg/m³, the volume deposited per day is about
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Convert mg/L to kg/m³, then mass to volume.
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Answer: B. 1,037 m³
2,000 mg/L = 2 kg/m³; trapped mass = 10 × 2 × 0.9 = 18 kg/s = 1.555 × 10⁶ kg/day; volume = 1.555 × 10⁶/1,500 = 1,037 m³/day.
111. The sediment storage zone of the above basin holds 1,200 m³. At 1,037 m³/day of deposition, flushing is required about every
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Divide storage volume by daily deposition.
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Answer: C. 28 hours
Time to fill = 1,200/1,037 = 1.16 days ≈ 27.8 hours, so the basin must be flushed roughly once a day.
112. In a continuously flushed settling basin (e.g. Dufour or Bieri type), the sediment is
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The plant need not be stopped.
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Answer: C. removed continuously through a bottom outlet with a small part of the flow, without stopping the basin
Continuous flushing systems use bottom slots or pipes with controlled discharge to remove deposits while the plant keeps running.
113. For effective hydraulic flushing of a settling basin, the basin floor should
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Water must be able to carry deposits out under gravity.
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Answer: C. slope towards the flushing outlet and the flushing gate should be large enough to create high velocity
Gravity flushing needs a longitudinal slope toward the outlet and sufficient flushing discharge velocity to scour the deposited sediment.
114. Sediment passing through the turbines of a high-head Pelton plant mainly causes
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Hard particles at high velocity erode metal.
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Answer: B. abrasive wear of the needle, nozzle and buckets
Hard particles such as quartz cut the nozzle, needle and buckets at high velocity, reducing efficiency and requiring repair, which is why settling basins are provided.
115. A turbine needs 10 m³/s. The intake and settling basin are designed for the turbine discharge plus a flushing allowance of 20%. The design discharge of the intake is
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Add the flushing allowance to the turbine flow.
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Answer: A. 12 m³/s
Design intake discharge = 10 × 1.20 = 12 m³/s.
116. A vortex tube sediment ejector or tunnel-type sediment excluder is used to
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They extract the bottom layer of flow.
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Answer: A. remove bed load from the flow before it reaches the headrace
These devices extract the bed layer of water that carries sediment and pass it back to the river through a tube or tunnel.
117. A hopper-type settling basin has
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The name describes the floor shape.
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Answer: D. sloping hoppers so deposits slide toward flushing pipes at the bottom
In a hopper basin, the floor is formed into inclined hoppers, from which the deposited sediment is drawn off through bottom outlets or pipes.
118. A free-flowing undersluice 3 m wide with a gate opening of 1.0 m works under an upstream head of 4 m. With Cd = 0.62 the discharge is about
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Gate-opening flow uses √(2gH).
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Answer: B. 16.5 m³/s
Q = Cd b a √(2gH) = 0.62 × 3 × 1.0 × √(78.48) = 0.62 × 3 × 8.859 = 16.5 m³/s.
119. According to the standard grain size scale, particles between 0.063 mm and 2 mm are classified as
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Think of the scale: clay, silt, sand, gravel.
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Answer: A. sand
Sand is 0.063-2 mm; silt is 0.002-0.063 mm, clay is finer than 0.002 mm and gravel is coarser than 2 mm. Settling basins mainly target sand-size particles.
120. A diversion weir at a run-of-river plant is built mainly to
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It has little storage.
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Answer: B. raise the water level so that the required discharge can enter the intake
The weir creates a small head pond and a controlled level, ensuring that the design flow can be diverted into the intake.
121. If the length and width of a settling basin are increased for the same discharge, the trap efficiency
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Think of Vetter's w·A/Q.
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Answer: D. increases because the surface area (and detention time) increases
A larger plan area reduces the surface loading Q/A and increases detention time, so more particles settle.
8.5 Water conveyance structures
31 questions · ACiE0805
122. A circular cross-section is generally preferred for pressure tunnels because
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Think of stress distribution under internal pressure.
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Answer: D. it resists internal water pressure with the most uniform stress distribution
In a circular section, internal pressure produces nearly uniform hoop stress and the section has the least perimeter for a given area, giving lower friction loss.
123. For a free-flow (non-pressure) headrace tunnel, the commonly used cross-section is
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A flat floor helps construction traffic.
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Answer: B. D-shaped or horseshoe, with a flat or nearly flat invert
D-shaped or horseshoe sections provide a flat floor for construction traffic and are well suited to free-surface flow and arch action in the roof.
124. A free-flow tunnel is not designed to flow completely full because
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Think of fluctuations and air.
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Answer: C. an air space is needed above the water to allow for surges and prevent pressure flow
Leaving a free board (air space) accommodates waves, surges and flow fluctuations and avoids unintended pressurisation of a non-pressure tunnel.
125. A circular concrete-lined tunnel of diameter 3.0 m (n = 0.014) flows full at a bed slope of 0.002. By Manning's equation, the discharge is about
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For a full circular pipe, R = D/4.
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Answer: B. 18.6 m³/s
R = D/4 = 0.75 m; V = (1/0.014) × 0.75^(2/3) × 0.002^0.5 = 2.64 m/s; Q = V × π(3²)/4 = 18.6 m³/s.
126. A pressure tunnel 2,500 m long and 3.5 m in diameter carries water at 2.8 m/s. With Darcy friction factor f = 0.015, the friction head loss is about
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Darcy-Weisbach equation.
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Answer: A. 4.28 m
hf = f L V²/(2 g D) = 0.015 × 2500 × 2.8²/(2 × 9.81 × 3.5) = 4.28 m.
127. The economic diameter of a headrace tunnel is the diameter that
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It balances cost against losses.
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Answer: A. minimises the sum of annual cost of the tunnel and the cost of energy lost in head loss
A larger tunnel costs more but loses less head; the economic size minimises total annual cost (construction plus value of lost energy).
128. In a headrace tunnel of a hydropower plant, the typical mean velocity adopted is of the order of
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Not too slow, otherwise the tunnel becomes huge.
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Answer: D. 2 to 4 m/s
Tunnel velocities of about 2-4 m/s balance the cost of excavation and lining against head loss and erosion.
129. The primary functions of a tunnel lining include all of the following EXCEPT
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Lining makes the surface smoother, not rougher.
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Answer: B. increasing the roughness of the tunnel surface
A lining smooths the surface (reduces friction), supports the rock, resists pressures and controls leakage; it does not increase roughness.
130. A steel liner in a pressure tunnel or shaft is used where
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Inadequate confinement requires strengthening.
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Answer: A. the internal pressure is greater than the rock confinement can resist
Steel liners resist high internal pressure and prevent hydrofracturing and leakage when the surrounding rock cover or strength is insufficient.
131. By the Norwegian confinement criterion, the minimum rock cover is C = γw h/(γr cos β). For a static head h = 120 m, rock unit weight 26.5 kN/m³, slope angle β = 30° and γw = 9.81 kN/m³, the required cover is about
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Include cos β in the denominator.
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Answer: A. 51 m
C = 120 × 9.81/(26.5 × cos 30°) = 1,177.2/22.95 = 51 m.
132. In Barton's Q-system, Q = (RQD/Jn)(Jr/Ja)(Jw/SRF). For RQD = 75, Jn = 9, Jr = 1.5, Ja = 1.0, Jw = 1.0 and SRF = 2.5, the Q value is
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Group into block size, joint strength and stress terms.
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Answer: B. 5.0
Q = (75/9) × (1.5/1.0) × (1.0/2.5) = 8.33 × 1.5 × 0.4 = 5.0 (fair rock).
133. An adit in a hydropower tunnel project is
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It is used for access during construction.
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Answer: A. a horizontal or inclined access tunnel driven to give additional working faces
Adits provide intermediate access to a long tunnel so that excavation can progress from several faces and the construction period is shortened.
134. A forebay in a run-of-river hydropower scheme is mainly provided to
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It is located at the junction of headrace and penstock.
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Answer: B. provide a transition between the headrace and penstock and a small storage to meet sudden load changes
The forebay provides a final settling, trash screening and a small volume to supply turbines during load changes while the headrace flow adjusts.
135. The minimum water level in a forebay is fixed so as to
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Think of a vortex draining a basin.
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Answer: A. prevent air entry (vortex formation) into the penstock
Adequate submergence over the penstock intake prevents vortices and entrainment of air that would harm the turbine and cause surging.
136. A surge tank in a hydropower system is provided mainly to
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It protects the tunnel from rapid pressure changes.
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Answer: B. reduce water hammer pressures in the penstock by partly reflecting the pressure waves
A surge tank acts as a free-surface reservoir near the turbine so that pressure waves from rapid valve movements do not travel in the long headrace tunnel.
137. A surge tank is usually placed
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Closer to the source of the disturbance.
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Answer: B. as close as possible to the powerhouse, at the end of the headrace tunnel
The tank should be near the turbines, so that the length of penstock exposed to water hammer is small, while the tunnel is protected by the tank.
138. In a differential (Johnson) surge tank, the riser is
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It has an inner and an outer part.
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Answer: A. an inner small-diameter tube that responds quickly, combined with an outer tank
A differential surge tank combines a riser and an annular tank: the riser quickly changes the pressure on load changes, and the outer tank provides storage.
139. The maximum upsurge (frictionless) in a simple surge tank on full load rejection is Z = V√(L At/(g As)). For V = 3 m/s, L = 1,500 m, tunnel area At = 9 m² and tank area As = 60 m², Z is about
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Take the square root of the whole ratio.
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Answer: B. 14.4 m
Z = 3 × √(1500 × 9/(9.81 × 60)) = 3 × √22.94 = 3 × 4.79 = 14.4 m.
140. Thoma's minimum stable area of a surge tank is As = L At/(2 g α H), with α = hf/V². For L = 3,000 m, At = 12 m², α = 0.96 s²/m, and H = 244 m, the Thoma area is about
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Substitute directly; mind the factor of 2 in the denominator.
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Answer: A. 7.8 m²
As = (3000 × 12)/(2 × 9.81 × 0.96 × 244) = 36,000/4,596 = 7.8 m². A tank larger than this is stable.
141. A penstock carries 6 m³/s at a mean velocity of 4 m/s. The required inside diameter is about
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Find the area first.
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Answer: D. 1.38 m
A = Q/V = 1.5 m²; D = √(4A/π) = √(1.91) = 1.38 m.
142. A steel penstock of 1.4 m diameter has a design head of 300 m (including surge). With an allowable stress of 140 MPa and a joint efficiency of 0.9, the thickness from t = pD/(2σe), ignoring corrosion allowance, is about
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Convert head to pressure in MPa first.
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Answer: C. 16.3 mm
p = 9.81 × 300 = 2.943 MPa; t = 2.943 × 1400/(2 × 140 × 0.9) = 16.3 mm.
143. In an exposed (surface) steel penstock, anchor blocks are provided mainly at
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Changes in direction create unbalanced forces.
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Answer: D. bends, to resist the unbalanced thrust and fix the pipe
Anchor blocks at bends and changes of slope resist unbalanced hydraulic and weight forces; expansion joints are placed between anchors.
144. The term 'pressure shaft' in a hydropower scheme refers to
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It carries high pressure down to the powerhouse.
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Answer: C. a steeply inclined or vertical high-pressure conduit in rock leading to the powerhouse
A pressure shaft is a steeply inclined or vertical conduit, usually steel-lined within concrete, that carries water at high pressure to the turbines.
145. Water hammer in a penstock is caused by
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It is a transient effect.
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Answer: D. a rapid change in velocity of flow, such as sudden closure of a valve or turbine gates
Rapid change in flow velocity converts kinetic energy into pressure waves, causing pressure rises and drops.
146. The pressure rise in a penstock due to instantaneous closure of a valve is given by Joukowsky's equation ΔH = aΔV/g. For wave velocity a = 1000 m/s and velocity change ΔV = 2.5 m/s, ΔH is about
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Divide by g, not multiply.
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Answer: C. 255 m
ΔH = 1000 × 2.5/9.81 = 255 m of water.
147. The pressure wave velocity in a steel pipe is a = 1/√(ρ(1/K + D/(Et))). For K = 2.07 GPa, D = 1.5 m, t = 20 mm, E = 210 GPa and ρ = 1000 kg/m³, a is about
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Pipe elasticity lowers the speed below that in unconfined water (~1,435 m/s).
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Answer: C. 1091 m/s
1/K = 4.83 × 10⁻¹⁰; D/(Et) = 3.57 × 10⁻¹⁰; sum = 8.40 × 10⁻¹⁰; a = 1/√(1000 × 8.40 × 10⁻¹⁰) = 1,091 m/s.
148. The critical time of closure for a pipe of length 600 m and wave velocity 1,200 m/s is
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The wave makes a round trip of 2L.
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Answer: D. 1.0 s
Tc = 2L/a = 2 × 600/1200 = 1.0 s. A closure faster than this is treated as rapid, giving the full Joukowsky rise.
149. For slow closure the approximate pressure rise is ΔH = 2LV/(gT). For L = 900 m, V = 3.5 m/s and a closure time T = 8 s, ΔH is about
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Substitute directly, keeping the factor 2.
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Answer: C. 80.3 m
ΔH = 2 × 900 × 3.5/(9.81 × 8) = 6,300/78.48 = 80.3 m.
150. In a Pelton turbine, a jet deflector is provided mainly to
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It lets the needle close slowly.
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Answer: D. avoid rapid closing of the needle and so limit water hammer in the penstock
On load rejection the deflector diverts the jet away from the runner, while the needle closes slowly, keeping pressure rise within limits.
151. To prevent sub-atmospheric pressure and possible collapse of a pressure tunnel or penstock, the conduit crown should lie
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Pressure must remain positive.
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Answer: C. below the minimum hydraulic gradient line throughout
If the crown rises above the hydraulic grade line, negative pressure develops, so the profile is kept below it with a margin.
152. Among the available cross-sections, the most economical rectangular channel (e.g. a headrace canal) has
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R = y/2.
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Answer: C. depth equal to half the width
For a rectangular section, minimum perimeter for a given area occurs when b = 2y, so the hydraulic radius is y/2.
8.6 Hydro-electric machines and powerhouse
32 questions · ACiE0806
153. Which of the following is an impulse turbine?
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Look for the free-jet machine.
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Answer: D. Pelton turbine
In an impulse turbine the whole head is converted to kinetic energy in nozzles, and the runner works at atmospheric pressure (Pelton). Francis, Kaplan and propeller are reaction turbines.
154. For a site with a net head of about 600 m and a small discharge of 4 m³/s, the most suitable turbine is
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High head, low flow.
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Answer: A. Pelton
High head and low discharge give a low specific speed, which suits Pelton turbines.
155. For a site with a net head of only 8 m and a large discharge, the most suitable turbine is
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Low head, high flow.
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Answer: D. Kaplan
Low head and large discharge give high specific speed, for which Kaplan (axial flow) turbines are best.
156. The specific speed Ns = N√P / H^(5/4) (N in rpm, P in kW, H in m) of a turbine running at 500 rpm, developing 10,000 kW under a net head of 100 m is about
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Evaluate H^1.25 = 316.2.
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Answer: C. 158
Ns = 500 × √10000 / 100^1.25 = 500 × 100/316.2 = 158.
157. The specific speed of a turbine is defined as the speed of a
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Unit power, unit head.
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Answer: A. geometrically similar turbine producing unit power under unit head
Specific speed is the speed of a homologous turbine that would develop one unit of power (1 kW or 1 hp) under one metre head.
158. Ordering turbines from the lowest to the highest specific speed gives
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Head decreases from Pelton to Kaplan.
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Answer: C. Pelton < Francis < Kaplan
Pelton turbines have the lowest Ns (about 10-60), Francis medium (about 60-400) and Kaplan/propeller the highest.
159. The jet velocity of a Pelton turbine with a nozzle coefficient of 0.98 under a net head of 400 m is about
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Torricelli velocity times coefficient.
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Answer: A. 86.8 m/s
V1 = Cv √(2gH) = 0.98 × √(2 × 9.81 × 400) = 0.98 × 88.59 = 86.8 m/s.
160. A Pelton wheel under 400 m head runs at 600 rpm with a speed ratio φ = 0.46. The pitch circle diameter is about
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u = πDN/60.
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Answer: B. 1.30 m
u = 0.46 × 88.59 = 40.75 m/s; D = 60u/(πN) = 1.30 m.
161. In an ideal Pelton wheel the maximum hydraulic efficiency (180° deflection, no friction) occurs when the bucket speed is
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Differentiate power with respect to u.
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Answer: D. half of the jet velocity
Power = ρQ(V − u)u(1 − cos θ) is maximum when u = V/2; in practice about 0.45 V is used because of friction.
162. A single-jet Pelton wheel has a specific speed of 20. If the same wheel is fitted with 4 jets, its specific speed becomes
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Power per jet falls as the number of jets rises.
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Answer: B. 40
Ns is proportional to √(number of jets), since each jet shares the power: Ns = 20 × √4 = 40.
163. In a Francis turbine, u1 = 30 m/s, Vw1 = 22 m/s, u2 = 15 m/s and Vw2 = 0. The Euler head (head extracted by the runner) is
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Euler's turbine equation.
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Answer: D. 67.3 m
He = (u1Vw1 − u2Vw2)/g = (30 × 22 − 0)/9.81 = 67.3 m.
164. The function of the draft tube in a reaction turbine is to
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It is the diffuser downstream of the runner.
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Answer: C. convert the kinetic energy at runner exit into pressure and allow the runner to be set above tailwater
A gradually expanding draft tube recovers velocity head and lets the runner discharge below atmospheric pressure, utilising the head between runner and tailwater.
165. The scroll (spiral) case of a Francis turbine has a cross-section that decreases along the flow direction in order to
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Flow is being drawn off continuously along its length.
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Answer: B. distribute water uniformly around the runner at nearly constant velocity
As water leaves the spiral casing through the stay vanes along its periphery, the casing area must reduce to keep the velocity of flow nearly constant.
166. In a Francis turbine, the discharge to the runner is regulated by
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Think of the movable vanes ahead of the runner.
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Answer: B. adjusting the opening of the wicket gates (guide vanes)
The governor turns the guide vanes, changing the flow area and flow rate and the power output.
167. A Kaplan turbine maintains good efficiency over a wide range of loads because
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Two things can be adjusted.
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Answer: D. both the runner blades and the guide vanes can be adjusted
A Kaplan turbine is double-regulated: blade pitch and guide vane opening can be matched to the flow.
168. Runaway speed of a turbine is the speed
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Think of load rejection without a working governor.
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Answer: B. reached at a given head when the load is thrown off and the governor fails to close the gates
Runaway speed (about 1.8 N for Pelton, 2-2.2 N for Francis, up to about 3 N for Kaplan) governs the mechanical design of the runner and generator.
169. The cavitation parameter (Thoma coefficient) of a turbine is defined as σ =
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NPSH-like numerator divided by net head.
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Answer: C. (Ha − Hv − Hs) / H
σ = (Ha − Hv − Hs)/H where Ha is atmospheric head, Hv vapour pressure head, Hs the suction (setting) height above tailwater and H the net head.
170. For a Francis turbine, Ha = 10.0 m, Hv = 0.25 m, net head 100 m and critical σ = 0.06. The maximum permissible setting of the runner above tailwater is
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Subtract the cavitation allowance σH.
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Answer: D. 3.75 m
Hs = Ha − Hv − σH = 10.0 − 0.25 − 0.06 × 100 = 3.75 m.
171. A hydro-generator with 12 poles is connected to a 50 Hz grid. Its synchronous speed is
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N = 120f/P.
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Answer: D. 500 rpm
N = 120 f / p = 120 × 50/12 = 500 rpm.
172. Hydro-generators coupled to slow-speed turbines are normally of
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Low speed needs many poles.
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Answer: A. salient-pole synchronous type with a large diameter and short axial length
Low speed requires many poles, which is accommodated by salient-pole construction with a large diameter; high-speed turbo-generators use cylindrical rotors.
173. A turbine delivers 18 MW to its generator. The generator efficiency is 97% and the rated power factor 0.90. The minimum generator rating is about
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Generators are rated in MVA, not MW.
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Answer: B. 19.4 MVA
Electrical output = 18 × 0.97 = 17.46 MW; rating = 17.46/0.90 = 19.4 MVA.
174. The governor of a hydro turbine controls
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It keeps the synchronous speed constant.
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Answer: A. the speed (frequency) by regulating the water flow to the turbine according to load changes
The governor senses speed deviations and moves the guide vanes or needle (and blades in a Kaplan) to match power with load, keeping frequency constant.
175. A pump delivers 0.2 m³/s against a total head of 60 m with an overall efficiency of 75%. The required shaft power is about
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Divide hydraulic power by efficiency.
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Answer: A. 157 kW
P = ρgQH/η = 9,810 × 0.2 × 60/0.75 = 156,960 W ≈ 157 kW.
176. A centrifugal pump develops a head of 40 m at 1,450 rpm. By the affinity laws, at 1,750 rpm the head is about
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Head varies with the square of speed.
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Answer: C. 58.3 m
H ∝ N²: H = 40 × (1750/1450)² = 58.3 m.
177. Two identical pumps are connected in parallel to a system. At the same head, the combined discharge is
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Parallel pumps add discharge.
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Answer: C. the sum of the discharges of the two pumps
In parallel operation the pumps share the same head and discharges are added; in series, heads add at the same discharge.
178. A pump draws water from a sump with a suction lift of 3.5 m. Atmospheric head 10.1 m, vapour pressure head 0.24 m and suction line losses 0.6 m. The available NPSH is
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Subtract all suction-side items from atmospheric head.
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Answer: A. 5.76 m
NPSHa = Ha − Hv − Hs − hf = 10.1 − 0.24 − 3.5 − 0.6 = 5.76 m; it must exceed the NPSH required to avoid cavitation.
179. An underground powerhouse is preferred over a surface powerhouse mainly when
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Good rock and unsafe surface conditions.
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Answer: A. the site has steep terrain with good rock and risks of landslides or avalanches on the surface
Where the valley sides are steep and unstable, or the penstock would be very long, a cavern powerhouse in sound rock is safer and often economical.
180. In a powerhouse, the overhead (EOT) crane is selected mainly according to
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Think of the largest item to be lifted.
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Answer: D. the heaviest single piece to be handled, usually the generator rotor or turbine runner
The crane must lift the heaviest assembled part (rotor or runner) during erection and maintenance.
181. The erection (service) bay of a powerhouse is provided for
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It is a floor area used for equipment handling.
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Answer: C. assembling and receiving equipment during installation and maintenance
The erection bay is an area at floor level, in the crane coverage, where machine parts are unloaded, assembled and repaired.
182. A powerhouse has 3 units spaced at 11 m, an erection bay of 14 m and 4 m additional space at the ends. Its overall length is
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Add unit bays, erection bay and end spaces.
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Answer: C. 51 m
Length = 3 × 11 + 14 + 4 = 51 m.
183. A plant has a net head of 120 m and a discharge of 30 m³/s. Turbine, generator and transformer efficiencies are 92%, 97% and 99.5%. The electrical output is about
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Multiply all three efficiencies.
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Answer: B. 31.4 MW
P = 9.81 × 30 × 120 × 0.92 × 0.97 × 0.995 = 31.4 MW (overall η = 0.888).
184. Which equipment isolates a turbine from the penstock for inspection and as a safety shut-off?
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It sits just upstream of the spiral case.
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Answer: B. Main inlet valve (butterfly or spherical valve)
A main inlet valve is placed ahead of the turbine, closed for maintenance or emergency; spherical valves are used at high head and butterfly valves at lower head.