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Chapter 3 · 8 hours

Hydro Power Plant Design

IOE past exam questions

Past questions and answers

32 questions set from this chapter, 11 of them more than once. Most asked first.

  • Asked 4 times
  • 2081 Shrawan · 6 marks
  • 2079 Shrawan · 6 marks
  • 2078 Kartik · 6 marks
  • 2075 Bhadra · 4 marks

Assume that you have identified a suitable hydropower site for electricity generation. Discuss the steps/stages how would you proceed to own, build that project and finally sell power to Nepal Electricity Authority.

Answer

To develop a hydropower project in Nepal as an Independent Power Producer (IPP), one must obtain licences from the Department of Electricity Development (DoED) under the Electricity Act 2049 (1992) and Electricity Regulation 2050, sign a Power Purchase Agreement (PPA) with NEA, and then finance, build and operate the plant.

Steps / stages

  1. Site identification and desk study: Check maps, hydrology data and existing licences to confirm the site is free.
  2. Company registration: Register a company at the Office of the Company Registrar (needed to hold licences).
  3. Survey licence: Apply to DoED for a survey licence (valid up to about 5 years) with a pre-feasibility report and fee. For large projects (above 200 MW) the Investment Board Nepal (IBN) is the approving body.
  4. Feasibility study: Hydrological, topographical, geological and sediment studies; optimise design discharge (e.g. Q40–Q45), head, installed capacity, layout; cost estimate and financial analysis.
  5. Environmental study: IEE or EIA as per the Environment Protection Act 2076 and Rules, with public hearing and approval from the concerned ministry.
  6. Grid connection: Apply to NEA for a connection agreement; NEA does a grid impact/connection study and fixes the substation and voltage of connection.
  7. Power Purchase Agreement (PPA): Sign the PPA with NEA (Power Trade Department) at the posted rates (separate wet- and dry-season rates per kWh with limited annual escalation), fixing the required commercial operation date (RCOD).
  8. Generation licence: Apply to DoED with the feasibility report, EIA/IEE approval, PPA and connection agreement; the licence is normally for up to 35 years, after which the project is handed to the government.
  9. Financial closure: Arrange debt (banks, typically 70%) and equity (promoters, about 30%); later issue shares to the public and to project-affected locals (IPO).
  10. Land and permits: Land acquisition, forest clearance, permits for explosives, roads.
  11. Construction: Tendering and contracts for civil, hydro-mechanical and electro-mechanical works and the transmission line to the NEA substation.
  12. Testing and commissioning: Wet tests, synchronisation and protection tests witnessed by NEA.
  13. Commercial operation (COD): Start selling energy to NEA as per the PPA; pay royalty to the government (shared with provinces and local levels).
  14. Operation and maintenance for the licence period, then transfer.
Site -> Survey licence -> Feasibility + EIA
     -> Connection agreement -> PPA (NEA)
     -> Generation licence -> Financial closure
     -> Construction -> COD -> Sell to NEA
  • Asked 2 times
  • 2082 Shrawan · 8 marks
  • 2074 Magh · 3+5 marks

List down the various components of a medium head relatively higher discharge hydropower plant and mention their functions.

Answer

A medium head (about 30–300 m), relatively high discharge plant is usually a run-of-river or pondage run-of-river scheme with a diversion weir/dam, a long large-diameter water conductor, a surge tank and a powerhouse with Francis turbines.

Layout

 Reservoir/headpond
  ____  Intake
 |    |__[TR]___ Headrace tunnel _________
 |Dam/|  (gate)                            \  Surge
 |weir|                                     \ tank
 |____|--> Spillway        Desander         [ST]
   |                                          |
   +-> Riparian release            Penstock   |
                                       \______|
                                        \
                                   [Powerhouse]
                                   Turbine+Gen
                                        |
                                    Tailrace -> River
                                        |
                              Switchyard -> Grid

Components and functions

A. Civil components

  1. Diversion weir / dam: Raises the water level and diverts flow to the intake; may create a small pondage.
  2. Spillway and gates: Pass flood water safely; undersluice gates flush sediment near the intake.
  3. Intake with trash rack: Draws the design discharge into the conveyance; the trash rack stops floating debris; gates control flow.
  4. Gravel trap and settling basin (desander): Remove sand and silt (particles above about 0.2 mm) to protect the turbines from abrasion — very important in Himalayan rivers.
  5. Headrace tunnel / canal: Carries the large discharge at a mild slope to the forebay/surge tank with low head loss.
  6. Surge tank (or forebay): Absorbs pressure rise (water hammer) during load rejection and supplies water during sudden load increase; protects the tunnel.
  7. Penstock / pressure shaft: Carries water under pressure from the surge tank to the turbines; ends in a manifold for each unit, with a valve.
  8. Powerhouse: Houses turbines, generators, control and auxiliary systems (surface or underground).
  9. Tailrace: Returns water from the draft tube to the river.

B. Mechanical components 10. Main inlet valve (butterfly/spherical): Isolates each turbine. 11. Francis turbine (suitable for medium head, high flow) with spiral casing, guide vanes and draft tube to recover kinetic energy. 12. Governor: Controls guide-vane opening to keep speed/frequency constant. 13. Cooling water, drainage, dewatering and EOT crane.

C. Electrical components 14. Synchronous generator with excitation system and AVR. 15. Generator circuit breaker, generator bus/leads, CTs and PTs. 16. Step-up (generator) transformer: Raises voltage (e.g. 11 kV to 132 kV). 17. Switchyard: Breakers, isolators, busbars, lightning arresters for connection to the transmission line. 18. Control, protection and SCADA; station auxiliary transformer; DC battery system; diesel generator for black start; earthing system.

  • Asked 2 times
  • 2081 Chaitra · 8 marks
  • 2073 Bhadra · 4+4 marks

Make a sketch to show various components of medium head and relatively large discharge, medium size Pondage Run of River (PRoR) hydropower plant. Also, mention the functions of each component.

Answer

A Pondage Run-of-River (PRoR) plant diverts river water through a water conductor like an ROR plant, but stores water in a pondage (daily storage) during off-peak hours so that the plant can run at full output during the peak hours (typically 4–6 hours in the evening). For medium head (about 50–300 m) and relatively large discharge, Francis turbines are used.

Sketch

        River
          |
   ======[Dam/Weir + Gates]======
   |  Pondage (daily storage)   |---> Spillway
   |_____________________________|
          |      |
          |      +--> Riparian release
       [Intake + trash rack]
          |
     [Desander / settling basin]
          |
   Headrace tunnel ==============\
                                  [Surge tank]
                                      |
                              Penstock |
                                      \|
                    [Powerhouse: MIV, Francis,
                     Generator, Governor, AVR]
                          |            |
                     Tailrace     GSU transformer
                          |            |
                       River      Switchyard -> Grid

Components and functions

ComponentFunction
Diversion dam / weir with gatesRaises water level, diverts water, forms pondage
PondageStores off-peak inflow for daily peaking operation
Spillway, undersluicePasses floods, flushes sediment
Intake with trash rackAdmits design flow; stops debris
Desander (settling basin)Removes fine sediment to protect turbines
Headrace tunnelConveys large discharge to the surge tank with low loss
Surge tankControls water hammer; supplies flow during load change
Penstock / pressure shaftCarries water under pressure to turbines
Main inlet valveIsolates each unit
Francis turbine + draft tubeConverts hydraulic energy to mechanical; draft tube recovers energy
GovernorControls speed/frequency and load
Synchronous generator + excitation/AVRGenerates power; controls voltage and reactive power
Generator CB, GSU transformerSwitching and step-up to transmission voltage
SwitchyardConnects plant to grid; isolation and protection
TailraceReturns water to the river
AuxiliariesStation transformer, DC system, diesel generator (black start), cooling, crane, SCADA

Why PRoR

It supplies the evening peak, which is valuable in Nepal, especially in the dry season (e.g. Kaligandaki A, Middle Marsyangdi are peaking ROR plants).

  • Asked 2 times
  • 2077 Chaitra · 8 marks
  • 2075 Bhadra · 6 marks

Draw a schematic diagram of PROR type hydropower plant showing all the necessary components and describe each of them.

Answer

A Peaking / Pondage Run-of-River (PROR) plant is a run-of-river scheme with a small daily storage (pondage) behind the diversion dam. Water is stored during low-demand hours and used to generate full power during peak hours. It combines the low cost of ROR with peaking capability.

Schematic

  River ->[ Diversion dam + radial gates ]-> Spillway
           |  Pondage (few hours storage) |
           +------------------------------+
                 |                 |
         [Intake + trash rack]   Riparian
                 |               release
        [Gravel trap / Desander]
                 |
       Headrace tunnel -------> [Surge tank]
                                     |
                              Penstock / shaft
                                     |
              +----------[ Powerhouse ]----------+
              | MIV -> Turbine -> Generator      |
              | Governor, Exciter, Control room  |
              +----------------------------------+
                 |                     |
              Tailrace           Transformer
                 |                     |
               River        Switchyard -> Line

Description of components

  1. Diversion dam/weir: Concrete or rockfill structure across the river that raises the water level and creates the pondage. Gates control water level.
  2. Pondage: Small reservoir that stores water for a few hours (e.g. 4–6 h) so that the plant can run at full capacity during the evening peak.
  3. Spillway and undersluice: Spillway passes flood flow safely; undersluice flushes sediment from the pondage.
  4. Intake: Admits the design discharge into the water conductor; has a trash rack to stop debris and gates for closing.
  5. Desander (settling basin): Settles sand and silt so that turbine runners are not eroded.
  6. Headrace tunnel/canal: Conveys water from intake to surge tank at a gentle slope.
  7. Surge tank: Vertical shaft that absorbs water hammer when turbines close suddenly and supplies water when load increases.
  8. Penstock/pressure shaft: Steel or lined pipe that carries water under high pressure to the turbines; branches into each unit.
  9. Powerhouse: Building housing main inlet valves, turbines (Francis/Pelton depending on head), generators, governors, excitation, control and protection equipment, cranes and auxiliaries.
  10. Turbine: Converts the energy of water into mechanical rotation.
  11. Generator: Synchronous machine converting mechanical power to electrical; excitation and AVR control voltage.
  12. Tailrace: Channel or tunnel that returns water to the river.
  13. Transformer and switchyard: Step up the generator voltage (e.g. 11 kV to 132 kV) and connect the plant to the transmission line with breakers, isolators and arresters.
  14. Auxiliary systems: Station service transformer, DC battery, diesel generator for black start, cooling water, fire protection, SCADA.

Examples in Nepal: Kaligandaki A (144 MW), Middle Marsyangdi (70 MW), Upper Tamakoshi (456 MW) are PROR plants.

  • Asked 2 times
  • 2074 Bhadra · 8 marks
  • 2071 Magh · 3+5 marks

Discuss the functions and importance of various components of hydropower plant with proper schematic diagram of ROR hydropower plant.

Answer

A run-of-river (ROR) plant diverts part of the river flow through a waterway to a powerhouse without any significant storage, so its output follows the natural river flow. Each component has one job: take water in, clean it, carry it with little head loss, convert its energy, and return it to the river.

Schematic of an ROR plant

 River
 ==|==  Diversion weir + intake (trash rack, gate)
   |
 [Gravel trap] -> [Desander / settling basin]
   |
   |  Headrace canal / tunnel (low slope)
   v
 [Forebay] or [Surge tank]
   |
   |  Penstock (steel pipe, steep)
   v
 [Valve]-[Turbine]-[Generator]  POWERHOUSE
   |              |
 Tailrace        GSU transformer -> Switchyard
   |              -> Transmission line
   v
 Back to river

Components, functions and importance

ComponentFunctionImportance
Diversion weir / barrageRaises water level and diverts flow into intakeGives a steady water level at intake
Intake with trash rack and gateAdmits the design discharge, stops floating debrisProtects the waterway; gate closes for floods and maintenance
Gravel trapRemoves coarse gravel just after intakePrevents canal choking
Desander (settling basin)Settles fine sediment (usually particles above 0.2 mm)Protects turbine runners from abrasion, very important in Himalayan rivers
Headrace canal / tunnelCarries water at low slope to the forebay or surge tankMost of the head is gained here, so head loss must be small
Forebay (canal system)Small pond before the penstock; final settling and spillProvides submergence and absorbs small load changes
Surge tank (tunnel system)Open shaft at end of pressure tunnelAbsorbs water hammer and supplies water on sudden load increase
PenstockPressure pipe from forebay/surge tank to turbineConverts elevation to pressure head; designed for water hammer
Main inlet valveButterfly or spherical valve before turbineIsolates the turbine for shutdown and emergency
TurbineConverts hydraulic energy to mechanical energyType chosen from head and discharge
GeneratorConverts mechanical energy to electrical energySynchronous machine with excitation and governor
Governor and excitationControl speed (frequency) and voltageKeep the unit stable and in step with the grid
Transformer and switchyardStep up voltage, switching and protectionConnect plant to the grid
TailraceReturns water to the riverMust avoid backwater that reduces net head

Key points

  • ROR plants are cheap and have small environmental impact, but output drops sharply in the dry season.
  • Because Nepalese rivers carry heavy sediment, the desander and good intake design decide the life of the turbines.
  • A riparian release (environmental flow) is always let past the weir for downstream life.
  • Asked 2 times
  • 2071 Bhadra · 2+6 marks
  • 2070 Bhadra · 3+5 marks

Discuss the various components of hydropower plant? Also mention their functions.

Answer

A hydropower plant converts the potential energy of water stored at a height into electrical energy. Its components fall into three groups: civil (to collect and carry water), mechanical (to convert water energy to shaft power) and electrical (to generate and deliver power).

 Reservoir/River
   | Dam or weir + intake
   v
 Headrace (canal/tunnel) -> Surge tank/forebay
   v
 Penstock -> Turbine -> Generator -> Transformer -> Grid
   |
 Draft tube -> Tailrace -> River

Civil components

  1. Dam or diversion weir: creates head and/or storage and diverts water into the intake.
  2. Intake: entry structure with trash rack and gates; controls flow and keeps out debris.
  3. Gravel trap and desander: remove coarse and fine sediment to protect the turbines.
  4. Headrace canal or tunnel: carries water to the forebay or surge tank with minimum head loss.
  5. Forebay / surge tank: forebay is a small pond feeding the penstock; surge tank absorbs water hammer pressure rise and supplies water on sudden load rise.
  6. Spillway: passes flood water safely over or around the dam.
  7. Powerhouse building: houses the units, cranes and control room.
  8. Tailrace: returns water to the river.

Mechanical components

  1. Penstock: steel pressure pipe carrying water to the turbine.
  2. Main inlet valve: butterfly (low/medium head) or spherical (high head) valve to isolate the turbine.
  3. Turbine: Pelton, Francis or Kaplan, converts hydraulic energy into rotating mechanical energy.
  4. Governor: controls wicket gates or needle to keep speed constant as load changes.
  5. Draft tube: (reaction turbines) recovers kinetic energy at runner exit and allows setting above tailwater.
  6. Auxiliaries: cooling water, oil pressure unit, compressed air, crane, drainage and dewatering.

Electrical components

  1. Generator: synchronous machine coupled to the turbine; produces power at e.g. 11 kV.
  2. Excitation system and AVR: supply DC to the rotor and control terminal voltage and reactive power.
  3. Generator circuit breaker and bus: switching and isolation of the unit.
  4. Generator step-up transformer: raises voltage (e.g. 11/132 kV) for transmission.
  5. Switchyard: breakers, isolators, CTs, PTs, lightning arresters and busbars.
  6. Station auxiliaries: station transformer, DC battery system, diesel generator for black start.
  7. Protection, control and SCADA: relays, metering and remote monitoring.
  • Asked 2 times
  • 2079 Jestha · 10 marks
  • 2072 Asoj · 8 marks

Sketch a typical layout of a medium head hydroelectric power station fitted with Francis turbine. Briefly describe the main elements.

Answer

A medium head station (roughly 30–300 m head) normally uses a Francis turbine, a mixed-flow reaction turbine. Water reaches the turbine under pressure through a long waterway and penstock, and leaves through a draft tube to the tailrace.

Typical layout

     Reservoir / pond
  ~~~~~~~~~~~~~~~~~~~~~
  |  Dam  | Intake + trash rack
  |       |====================\  Headrace tunnel
  |_______|                     \
                            [Surge tank]
                                 |
                                 | Penstock
                                 |
               Valve --> +-------v------+
                         | Spiral casing|
                         |  Guide vanes |
                         |  Francis     |
                         |  runner      |---Generator
                         +------+-------+
                                | Draft tube
                                v
                          Tailrace --> River

Main elements

  1. Reservoir or pondage and dam: stores water and creates part of the head. A spillway passes floods.
  2. Intake with trash rack and gates: admits water and keeps out debris; gates allow the waterway to be dewatered.
  3. Headrace tunnel / pressure conduit: carries water to the surge tank with a gentle slope; it is often lined to reduce friction loss.
  4. Surge tank: an open vertical shaft at the end of the tunnel. When the turbine gates close suddenly, the rising water level in the tank absorbs the water hammer; when load increases, it supplies water until the tunnel flow speeds up. It keeps water hammer confined to the short penstock.
  5. Penstock: steel pipe from the surge tank to the powerhouse; designed for static head plus water hammer. Anchor blocks and expansion joints are provided.
  6. Main inlet valve: usually a butterfly valve at medium head; closes in emergency.
  7. Spiral (scroll) casing: distributes water evenly around the runner; its area decreases along the flow so the velocity stays uniform.
  8. Stay vanes and guide vanes (wicket gates): stay vanes support the casing; guide vanes, moved by the governor, control the flow and direct it onto the runner at the correct angle.
  9. Francis runner: water enters radially and leaves axially; both pressure and kinetic energy are converted into torque. Medium specific speed (about 60–300, metric, P in kW).
  10. Draft tube: a diverging tube from runner exit to tailwater. It recovers kinetic energy and creates suction head so the runner can be set above tailwater; its setting is limited by cavitation (Thoma's coefficient).
  11. Generator: vertical-shaft synchronous generator directly coupled to the runner; its speed is fixed by N=120f/pN = 120f/p.
  12. Governor and excitation system: control speed (frequency) and voltage.
  13. Powerhouse: surface or underground; contains units, EOT crane, control room and auxiliaries.
  14. Tailrace: channel that returns water to the river.
  15. Transformer and switchyard: step up the voltage and connect to the transmission line.

Example

Kaligandaki "A" (144 MW, about 115 m net head) and Middle Marsyangdi (70 MW) in Nepal use vertical Francis units with this type of layout.

  • Asked 2 times
  • 2079 Shrawan · 4 marks
  • 2077 Chaitra · 4 marks

What is scientific method of turbine selection?

Answer

The scientific method of turbine selection chooses the turbine type from the specific speed (NsN_s) that the site's head, discharge and unit speed demand, and then checks it against the head range, efficiency curve and cavitation limits, rather than choosing by habit.

Steps

  1. Find net head and design discharge per unit from the flow duration curve (e.g. Q45Q_{45}), after deducting riparian release and head losses.
  2. Find unit power:
P=ηtρgQHnet (W)P = \eta_t \rho g Q H_{net}\ \text{(W)}
  1. Choose synchronous speed for the generator: N=120fpN = \dfrac{120 f}{p} (e.g. 500, 600, 750, 1000 rpm at 50 Hz).
  2. Compute specific speed:
Ns=NPH5/4(N in rpm, P in kW, H in m)N_s = \frac{N\sqrt{P}}{H^{5/4}}\quad (N\ \text{in rpm},\ P\ \text{in kW},\ H\ \text{in m})
  1. Pick the turbine type whose range contains NsN_s:
TurbineNsN_s (metric, kW)Usual head
Pelton10–35 per jet50–1300 m
Francis60–30010–350 m
Kaplan / propeller300–10002–40 m
  1. Check with the manufacturer's head–discharge application chart, part-load efficiency, and cavitation (Thoma's σ\sigma and setting level).
  2. Adjust number of units or speed if NsN_s falls in an overlap zone, and compare cost.
  • Asked 2 times
  • 2081 Shrawan · 4 marks
  • 2074 Bhadra · 5 marks

Mention major factors governing during appropriate turbine selection.

Answer

The turbine type is selected mainly from the head and discharge of the site, then refined by speed, efficiency and site conditions.

Major governing factors

  1. Net head: the most important factor. High head (above ~300 m) suits Pelton; medium head (30–300 m) Francis; low head (below ~40 m) Kaplan/propeller.
  2. Discharge and its variation: large discharge at low head favours Kaplan; small discharge at high head favours Pelton. A river with widely varying flow needs a turbine with a flat part-load efficiency curve.
  3. Specific speed: Ns=NP/H5/4N_s = N\sqrt{P}/H^{5/4} combines head, power and speed into one number that fixes the runner type.
  4. Rotational speed: higher speed means a smaller, cheaper generator, but is limited by cavitation and runaway speed.
  5. Part-load efficiency: Pelton and Kaplan keep high efficiency at part load; Francis efficiency falls quickly below about 40–50% load.
  6. Cavitation and setting level: reaction turbines must be set low enough (Thoma's σ\sigma), which adds excavation cost.
  7. Sediment in water: silty Himalayan rivers wear Francis runners quickly; Pelton is easier to repair.
  8. Cost, size and transport: runner size, powerhouse size and access road limits.
  9. Overall efficiency and maintenance needs, and availability of spares and local expertise.
  • Asked 2 times
  • 2073 Bhadra · 6 marks
  • 2071 Magh · 6 marks

How the 'specific speed' is defined for a turbine? Discuss its purpose.

Answer

Specific speed (NsN_s) of a turbine is the speed at which a geometrically similar model turbine would run if it produced 1 kW of power under a head of 1 m. It is a type number that describes the shape of the runner.

Derivation (outline)

For similar turbines, the jet/flow velocity varies as V∝HV \propto \sqrt{H} and the peripheral speed u=πDN/60∝Hu = \pi D N/60 \propto \sqrt{H}, so D∝H/ND \propto \sqrt{H}/N. Discharge Q∝D2HQ \propto D^2\sqrt{H} and power P∝QH∝D2H3/2P \propto QH \propto D^2 H^{3/2}. Substituting DD:

P∝HN2 H3/2=H5/2N2⇒N=K H5/4P\begin{aligned} P &\propto \frac{H}{N^2}\,H^{3/2} = \frac{H^{5/2}}{N^2} \\ \Rightarrow N &= K\,\frac{H^{5/4}}{\sqrt{P}} \end{aligned}

When P=1P = 1 kW and H=1H = 1 m, N=K=NsN = K = N_s. Hence

Ns=NPH5/4N_s = \frac{N\sqrt{P}}{H^{5/4}}

with NN in rpm, PP in kW and HH in m.

Purpose

  1. Selection of turbine type: each type works best in a band of NsN_s:
TurbineNsN_s range (kW, m)
Pelton (single jet)10–35
Pelton (multi jet)35–60
Francis60–300
Kaplan / propeller300–1000
  1. Comparison of turbines: turbines of different size are compared on one scale.
  2. Model testing: a model and prototype with the same NsN_s are geometrically and dynamically similar, so model results can be scaled up.
  3. Fixing speed and unit size: for a given head and power, choosing NsN_s fixes the rotational speed, and hence the generator poles and cost.
  4. Cavitation check: the Thoma critical σ\sigma rises with NsN_s, so NsN_s fixes how low the runner must be set.

Example: N=600N = 600 rpm, P=5000P = 5000 kW, H=200H = 200 m gives Ns=6005000/2001.25≈56.4N_s = 600\sqrt{5000}/200^{1.25} \approx 56.4, which lies at the boundary of multi-jet Pelton and slow Francis, so both would be compared.

  • Asked 2 times
  • 2080 Chaitra · 4 marks
  • 2078 Chaitra · 4 marks

What is flow duration curve? Also discuss about its applicability.

Answer

A flow duration curve (FDC) is a plot of river discharge against the percentage of time that discharge is equalled or exceeded. It is drawn by arranging recorded (daily or monthly) flows in descending order and plotting each flow against its exceedance probability, P=mn+1×100%P = \dfrac{m}{n+1}\times 100\% (m = rank, n = number of values).

 Q (m3/s)
  |\
  | \
  |  \__
  |     \____
  |          \______
  |                 \_______
  +--------------------------> % time exceeded
  0     45  65        95   100

Applicability

  1. Design discharge: the turbine flow is chosen at a set exceedance, e.g. Q40Q_{40}–Q45Q_{45} for ROR projects in Nepal.
  2. Installed capacity and energy: the area under the FDC (limited by the design flow) gives the energy available; it is used to compute annual energy and plant factor.
  3. Firm power: flow at 90–95% exceedance gives the firm (dependable) power.
  4. Number and size of units: the shape of the curve shows how long the plant will run at part load, so it guides unit sizing.
  5. Riparian release and water rights: low-flow end of the curve shows the minimum flow to release downstream.
  6. Comparison of sites: a flat curve means steady flow (good for ROR); a steep curve means highly variable flow needing storage.
  • 2079 Shrawan · 8 marks

Stating environmental impact of P-ROR type hydro-power plant, draw a schematic diagram showing all the necessary components and describe each of them.

Answer

A peaking run-of-river (P-ROR) plant is an ROR plant with a small daily pondage, so that water collected during off-peak hours is used to run the plant at full capacity during the evening peak (typically 4–6 hours).

Environmental impacts of P-ROR

  • Hydro-peaking downstream: flow below the tailrace rises and falls sharply every day; this disturbs fish, aquatic insects and people using the riverbank.
  • Dewatered stretch: between the dam and tailrace only the riparian release flows, reducing habitat, fish migration and water for irrigation/water mills.
  • Pondage submergence: a small area of land and forest is flooded; some houses or farmland may be affected.
  • Sediment trapping and flushing: the pond traps sediment; periodic flushing releases very turbid water downstream.
  • Construction impacts: spoil from tunnels, access roads, landslides, dust, noise and inflow of workers.
  • Positive impacts: clean peaking energy that replaces diesel/thermal imports, local roads, jobs and electrification.

Mitigation: adequate riparian release (at least 10% of minimum monthly flow in Nepal), fish ladders, controlled flushing, re-afforestation and spoil management.

Schematic of a P-ROR plant

 River
 ==|== Dam/weir with spillway + undersluice
   |
 Intake (trash rack, gate)
   |
 Gravel trap -> Desander
   |
 [ DAILY PONDAGE ] (stores off-peak flow)
   |
 Headrace tunnel
   |
 Surge tank
   |
 Penstock (+ valve house)
   |
 Powerhouse: Turbine-Generator-Transformer
   |                       |
 Tailrace -> River     Switchyard -> Grid

Components

  1. Diversion dam/weir: raises water level; gated spillway passes floods; undersluice flushes sediment at the intake.
  2. Intake: trash rack and gates admit the design discharge.
  3. Gravel trap and desander: remove sediment (usually >0.2 mm) before water enters the pondage and tunnel.
  4. Daily pondage: a reservoir sized to store the inflow of off-peak hours so the plant can run at full output during peak hours. Its volume is about (Qd−Qdry)×tpeak(Q_d - Q_{dry})\times t_{peak}.
  5. Headrace tunnel: pressure tunnel carrying water to the surge tank.
  6. Surge tank: absorbs water hammer and meets sudden load rise.
  7. Penstock and valve house: steel pipe to the powerhouse with an emergency butterfly valve at the top.
  8. Powerhouse: turbines (Francis or Pelton depending on head), generators, governors, excitation, control and auxiliaries.
  9. Tailrace: returns water to the river; may include a re-regulating pond to reduce hydro-peaking.
  10. Switchyard and transmission line: step up and evacuate power to the grid.
  • 2078 Kartik · 8 marks

Discuss major components of a Daily Pondage Run off the River type hydropower project with suitable schematic diagram.

Answer

A daily pondage run-of-river (PROR) project stores river water for a few hours each day in a small pond so the plant can generate more during the daily peak load. It has all ROR components plus the pondage.

Schematic

 River ==|== Weir/dam + spillway + undersluice
         |
       Intake ---> Gravel trap ---> Desander
                                       |
                              [ DAILY PONDAGE ]
                                       |
                              Headrace tunnel/canal
                                       |
                                 Surge tank/forebay
                                       |
                                    Penstock
                                       |
                         Valve -> Turbine -> Generator
                                       |        |
                                   Tailrace   Transformer
                                       |        |
                                     River   Switchyard

Major components

  1. Headworks: diversion weir or low dam to raise water level; gated spillway for floods; undersluice to flush sediment away from the intake.
  2. Intake: side or frontal intake with trash rack and gates to admit design discharge.
  3. Gravel trap: removes gravel and coarse sand near the intake.
  4. Desander (settling basin): long, slow-flow basin where fine sediment settles; flushed periodically. Essential in Nepal to protect turbines.
  5. Daily pondage: small reservoir (or enlarged desander/forebay) that stores the off-peak inflow. Storage needed ≈(Qdesign−Qinflow)×\approx (Q_{design} - Q_{inflow}) \times peak hours. It lets a dry-season plant give full output for about 4–6 peak hours.
  6. Headrace tunnel or canal: carries water to the surge tank/forebay with small head loss.
  7. Surge tank (or forebay): absorbs pressure surges and supplies water during sudden load increase.
  8. Penstock: steel pressure pipe with anchor blocks; valve at the top for emergency closure.
  9. Powerhouse: turbine (Francis for medium head, Pelton for high head), generator, governor, excitation, cooling and control systems.
  10. Tailrace: returns water to the river.
  11. Electrical works: step-up transformer, switchyard and transmission line to the grid.

Advantages

Higher peak capacity and revenue than plain ROR, better match with Nepal's evening peak, and much smaller submergence than a storage project. Examples in Nepal: Kaligandaki "A", Middle Marsyangdi and Upper Tamakoshi are peaking ROR projects.

  • 2078 Chaitra · 8 marks

Discuss major components of a Daily Pondage Run off the River type hydropower project with tentative head 150 m and discharge 10 m³/s per unit. Illustrate with suitable schematic diagram.

Answer

For a head of about 150 m and 10 m³/s per unit, the unit output is roughly

P≈9.81×10×150×0.9≈13.2 MW (shaft)P \approx 9.81 \times 10 \times 150 \times 0.9 \approx 13.2\ \text{MW (shaft)}

and the specific speed at 600 rpm is Ns=60013244/1505/4≈132N_s = 600\sqrt{13244}/150^{5/4} \approx 132, which lies well inside the Francis range (60–300). So the project is a medium-head daily pondage run-of-river (PROR) scheme with vertical Francis units of roughly 12–13 MW each.

Schematic

 River ==|== Diversion weir + gated spillway
         |
      Intake (trash rack, gates)
         |
      Gravel trap -> Desander (2 or more bays)
         |
     [ DAILY PONDAGE ] -- spillway/flushing
         |
      Headrace tunnel (pressurised)
         |
      Surge tank
         |
      Penstock (steel, ~150 m head)
         |
      Butterfly valve
         |
      Francis turbine + Generator (~13 MW)
         |                 |
      Draft tube        Unit transformer
         |                 |
      Tailrace -> River  Switchyard -> Grid

Major components

  1. Diversion weir and spillway: raise water level and pass monsoon floods; undersluice keeps the intake free of sediment.
  2. Intake: sized for total design discharge (10 m³/s × number of units) plus flushing water; trash rack and gates.
  3. Gravel trap and desander: remove particles larger than about 0.2 mm, as sediment at 150 m head quickly erodes Francis runners and guide vanes.
  4. Daily pondage: stores off-peak inflow; for example, to run a 10 m³/s unit for 5 peak hours when the dry-season inflow is only 5 m³/s, storage =(10−5)×5×3600=90,000= (10-5)\times 5 \times 3600 = 90{,}000 m³ is needed.
  5. Headrace tunnel: carries water under low pressure to the surge tank.
  6. Surge tank: absorbs water hammer when governor closes the wicket gates and supplies water on load increase.
  7. Penstock: about 150 m static head plus water hammer allowance (typically 20–30%), with anchor blocks and expansion joints.
  8. Main inlet valve: butterfly valve, suitable for medium head.
  9. Francis turbine: spiral casing, stay vanes, wicket gates, runner and draft tube; vertical shaft; runner set below tailwater as required by cavitation.
  10. Generator: vertical synchronous generator at 11 kV, e.g. about 15 MVA at 0.85 pf, with excitation system and governor.
  11. Tailrace: returns water to the river below the powerhouse.
  12. Electrical system: unit step-up transformer (11/132 kV), switchyard, station supply and diesel generator for black start.
  • 2070 Magh · 8 marks

Compare the components, operation and characteristics of run of river (ROR) type hydropower plant with pondage run of river (PROR).

Answer

A run-of-river (ROR) plant uses the river flow as it comes, with no storage. A pondage run-of-river (PROR) plant adds a small pond that stores water for a few hours so the plant can follow the daily load and give more power at peak time.

AspectRORPROR
StorageNone (only forebay)Daily pondage, a few hours of storage
HeadworksLow weirWeir or slightly higher dam to create pond
Extra component—Pondage with flushing arrangement
OperationBase load; output follows river flowPeaking; stores off-peak water, generates more at peak
Dry-season outputLow, nearly constantSame energy, but concentrated into peak hours
Installed capacityBased on design flow (e.g. Q45Q_{45})Can be higher than ROR for same river
Load followingPoorGood, within the pond volume
CostLowestHigher (pond, larger dam, gates)
Environmental impactLeast; dewatered stretch onlyAdds small submergence and hydro-peaking downstream
SedimentPasses through after desanderPond traps sediment, needs flushing
Value to gridEnergy onlyEnergy plus peaking capacity

Components

Both have: weir/dam, intake, gravel trap, desander, headrace, forebay or surge tank, penstock, powerhouse, tailrace and switchyard. PROR adds the pondage (and its spillway and flushing gates) and usually has a pressure tunnel with surge tank.

Operation and characteristics

  • ROR: turbine discharge equals river inflow minus riparian release; in the wet season it runs at full load and spills the excess, in the dry season it runs at part load.
  • PROR: during the night the pond fills; in the 4–6 peak hours the plant runs at full load using inflow plus stored water. Total daily energy is about the same as ROR, but its value is higher because peak energy is scarce in Nepal.
  • Examples: many small IPP plants in Nepal are ROR; Kaligandaki "A" and Middle Marsyangdi are PROR.
  • 2080 Chaitra · 8 marks

Draw schematic diagram of a hydropower plant and explain the basic civil, mechanical and electrical components used.

Answer

A hydropower plant converts the potential energy of water into electricity. Its works are grouped into civil, mechanical and electrical components.

 River/Reservoir
  ==|== Dam/weir             [CIVIL]
     |  Intake, desander
     |  Headrace tunnel/canal
     |  Surge tank / forebay
     v
  Penstock -> Valve          [MECH]
     v
  Turbine ---- Generator     [MECH] [ELEC]
     |  Governor  |  Excitation
  Draft tube      v
     |        GCB -> GSU transformer
  Tailrace        v           [ELEC]
     v        Switchyard -> Transmission line
   River

Civil components

  • Dam / diversion weir: raises water level, creates head or storage; includes spillway and undersluice.
  • Intake: trash rack and gates admit design flow.
  • Gravel trap and desander: remove sediment.
  • Headrace canal or tunnel: conveys water with low head loss.
  • Forebay or surge tank: provides a free water surface, absorbs water hammer.
  • Powerhouse: building or cavern for units, crane and control room.
  • Tailrace: returns water to the river.
  • Access roads, anchor blocks and saddles for the penstock.

Mechanical components

  • Penstock: steel pipe that delivers water under pressure.
  • Main inlet valve: butterfly or spherical valve.
  • Turbine: Pelton (high head), Francis (medium head), Kaplan (low head).
  • Governor: controls flow to keep speed and frequency constant.
  • Draft tube: recovers outlet energy of reaction turbines.
  • Auxiliaries: cooling water, lubrication and oil pressure units, compressed air, EOT crane, dewatering and drainage pumps, fire fighting.

Electrical components

  • Generator: synchronous, usually 11 kV.
  • Excitation system and AVR: brushless or static; control voltage and reactive power.
  • Generator circuit breaker and bus duct.
  • Generator step-up transformer: e.g. 11/132 kV.
  • Switchyard: circuit breakers, isolators, CTs, PTs, lightning arresters, busbars.
  • Auxiliary supply: station transformer, DC battery and charger, diesel generator for black start.
  • Protection, control, metering, SCADA and communication (PLCC/OPGW).
  • Earthing system: grounding mat for powerhouse and switchyard.
  • 2079 Chaitra · 8 marks

Draw a schematic diagram of reservoir type hydro-power plant and describe necessary civil, mechanical and electrical components.

Answer

A reservoir (storage) type hydropower plant has a high dam that stores water in a large reservoir, usually for seasonal use. Water stored in the wet season is released in the dry season, so the plant can supply firm and peaking power all year. Kulekhani-I (60 MW) is Nepal's main example.

Schematic

  Reservoir (FSL)  ~~~~~~~~~~~~
  ~~~~~~~~~~~~~~~ |\
  ~~~ storage ~~~ | \  High dam
  ~~~~~~~~~~~~~~~ |  \  + spillway
  Intake tower -->|===\=============\
  (MOL)           |____\             \ Headrace tunnel
                                 [Surge tank]
                                      |
                                      | Penstock
                                      v
                    Valve -> Turbine -> Generator
                               |          |
                          Draft tube   Transformer
                               |          |
                           Tailrace   Switchyard
                               v          v
                             River       Grid

Civil components

  1. Dam: concrete gravity, arch or rockfill dam that creates head and storage. Full supply level (FSL) and minimum operating level (MOL) define the live storage.
  2. Reservoir: stores water from the monsoon for the dry season; dead storage below MOL holds sediment.
  3. Spillway: gated or ungated, passes the design flood safely.
  4. Intake tower: draws water from various levels with trash rack and gates.
  5. Headrace tunnel: pressure tunnel to the surge tank.
  6. Surge tank: protects the tunnel from water hammer and supplies water on load rise.
  7. Diversion tunnel/cofferdam (during construction), bottom outlet for emptying and flushing.
  8. Powerhouse (surface or underground) and tailrace.

Mechanical components

  1. Penstock with anchor blocks; main inlet valve (butterfly or spherical).
  2. Turbine: Francis for medium head or Pelton for high head (Kulekhani uses Pelton).
  3. Governor: keeps frequency constant by controlling flow.
  4. Draft tube (reaction turbines), cooling water, oil and compressed air systems, EOT crane.

Electrical components

  1. Synchronous generator with excitation system and AVR.
  2. Generator circuit breaker and bus duct.
  3. Step-up transformer and switchyard (breakers, isolators, CT, PT, lightning arresters).
  4. Station service: station transformer, DC battery, diesel generator for black start.
  5. Protection, control and SCADA; earthing mat.

Features

Reservoir plants give firm power, peaking and frequency regulation, but have high cost, long construction time, resettlement and sedimentation problems.

  • 2081 Shrawan · 6 marks

Discuss major components of a high dam type hydropower project with suitable schematic diagram.

Answer

A high dam type project uses a tall dam (usually above about 15 m, often 100 m or more) to create a large head and storage reservoir; the powerhouse is placed at the toe of the dam or at the end of a short waterway.

 Reservoir  ~~~~~~~~~~|\
 ~~~~~~~~~~~~~~~~~~~~~| \  High dam
 Intake (trash rack)->|==\====\
                      |   \    \ Penstock
                      |____\    \
                     Spillway  [Powerhouse]
                                Turbine-Gen
                                   |    |
                            Tailrace  Transformer
                                   |    |
                                River  Switchyard

Major components

  1. High dam: concrete gravity, arch or rockfill/earthfill; creates head and storage. It must be safe against overturning, sliding and seepage.
  2. Reservoir: live storage between full supply level and minimum operating level; dead storage for sediment.
  3. Spillway and energy dissipator: pass floods safely and protect the riverbed below.
  4. Intake: at the dam face or in a tower, with trash rack and gates, placed above sediment level.
  5. Bottom outlet / sluice: to flush sediment and empty the reservoir.
  6. Penstock: short pressure pipe through or around the dam to the turbines (surge tank needed only if the waterway is long).
  7. Powerhouse: at the dam toe; contains turbines (usually Francis), generators, governors and auxiliaries.
  8. Tailrace: returns water to the river.
  9. Electrical works: step-up transformer, switchyard and transmission line.

Example: the proposed Budhigandaki (about 1200 MW) and Kulekhani reservoir in Nepal; worldwide, Three Gorges and Tehri.

Advantages and limitations

  • Gives large head with a short waterway, firm power through the dry season, peaking capacity and flood control.
  • Needs very high investment and long construction time, causes submergence and resettlement, and loses storage to sedimentation, which is serious in Himalayan rivers.
  • 2072 Magh · 8 marks

Draw the schematic diagram of hydropower plant showing different structural components. State the role of surge tank in hydropower plant construction.

Answer

The structural (mainly civil) components of a hydropower plant collect, store, clean and carry water to the turbine and return it to the river.

Schematic

 Reservoir/River
  ==|== Dam/weir + spillway
     |
   Intake (trash rack, gates)
     |
   Desander
     |
   Headrace tunnel ================\
                                   |
                            [SURGE TANK]
                                   |
                              Penstock
                                   |
                        [POWERHOUSE: turbine,
                         generator]
                                   |
                              Tailrace -> River

Structural components

  1. Dam or weir with spillway: creates head/storage and passes floods.
  2. Intake: admits water; trash rack keeps out debris.
  3. Gravel trap and desander: remove sediment.
  4. Headrace canal or tunnel: carries water at a gentle slope.
  5. Forebay or surge tank: free water surface at the head of the penstock.
  6. Penstock with anchor blocks and saddle supports.
  7. Powerhouse: foundation, superstructure, crane beams.
  8. Tailrace channel.

Role of surge tank

A surge tank is an open vertical shaft or chamber placed at the junction of the long low-pressure headrace tunnel and the steep penstock.

  • Protects against water hammer: when load is rejected, the governor closes the gates quickly. The moving water column cannot stop instantly; without a surge tank the pressure rise would travel through the whole tunnel. The surge tank lets the water rise in it, so the high pressure is limited to the short penstock.
  • Supplies water on load increase: when gates open suddenly, water is drawn from the tank until the slow tunnel flow speeds up, avoiding a drop in pressure (and vacuum) in the penstock.
  • Reduces penstock and tunnel thickness: lower design pressure means cheaper tunnel lining.
  • Improves governing: reduces the effective water inertia time constant Tw=LVgHT_w = \dfrac{LV}{gH}, since only the penstock length counts, making speed regulation stable.
  • Types: simple, restricted orifice, differential and chamber types.
  • 2078 Kartik · 4+4 marks

What is scientific method of turbine selection? Mention major factors governing turbine during selection of appropriate turbine type.

Answer

Scientific method of turbine selection

Turbine selection by the scientific method is based on specific speed and the site's head–discharge data, checked against standard application charts.

  1. Find net head HH and design discharge per unit QQ (e.g. Q45Q_{45} from the FDC less riparian release).
  2. Find unit output P=ηtρgQHP = \eta_t \rho g Q H.
  3. Choose a synchronous speed N=120f/pN = 120f/p.
  4. Calculate
Ns=NPH5/4N_s = \frac{N\sqrt{P}}{H^{5/4}}

(N in rpm, P in kW, H in m). 5. Select the type from the NsN_s range: Pelton 10–35 (single jet), Francis 60–300, Kaplan 300–1000. 6. Cross-check with the head range (Pelton 50–1300 m, Francis 10–350 m, Kaplan 2–40 m) and the manufacturer's HH–QQ chart; check cavitation (Thoma's σ\sigma) and part-load efficiency.

Major factors governing turbine selection

  1. Net head: primary factor; fixes the broad type.
  2. Discharge and its variation: low head–high flow suits Kaplan; high head–low flow suits Pelton.
  3. Specific speed and rotational speed: higher speed gives smaller generator but more cavitation risk.
  4. Part-load efficiency: Pelton and Kaplan have flat efficiency curves; Francis drops at low load.
  5. Cavitation and setting: reaction turbines need submergence, adding excavation cost.
  6. Sediment content: high silt erodes Francis runners; Pelton runners are easier to repair.
  7. Number of units and unit size, transport limits.
  8. Cost, maintenance and runaway speed.

Example

Net head 300 m, design flow 4 m³/s for one unit: P≈9.81×4×300×0.9≈10.6P \approx 9.81\times 4\times 300\times 0.9 \approx 10.6 MW. At 600 rpm, Ns=60010595/3005/4≈49N_s = 600\sqrt{10595}/300^{5/4} \approx 49, which is in the multi-jet Pelton range; a 2-jet or 4-jet Pelton (or a slow Francis) would be compared on cost and part-load efficiency.

  • 2072 Magh · 8 marks

What do you mean by discharge exceedance? Describe the criterion for selecting turbine in hydro-electric power plant.

Answer

Discharge exceedance

Discharge exceedance is the percentage of time a given river discharge is equalled or exceeded. It is read from the flow duration curve (FDC). For example, Q45=10Q_{45} = 10 m³/s means the river flow is 10 m³/s or more for 45% of the time (about 164 days a year).

  • Low exceedance (e.g. Q25Q_{25}) → large flow, available for a short time.
  • High exceedance (e.g. Q95Q_{95}) → small but dependable (firm) flow.
  • In Nepal, ROR projects are usually designed at about Q40Q_{40}–Q45Q_{45}; firm power is judged from Q90Q_{90}–Q95Q_{95}.
  • Exceedance is found by ranking the recorded flows in descending order and computing P=mn+1×100%P = \dfrac{m}{n+1}\times 100\% for rank mm out of nn values.
  • The choice of design exceedance is a trade-off: a lower exceedance gives more installed capacity and energy but a lower plant factor and higher cost per kW.
 Q
 |\
 | \
 |  \___ Q45
 |      \_____ Q65
 |            \______ Q95
 +---------------------> % exceedance

Criteria for selecting turbine

  1. Net head: Pelton for high head (50–1300 m), Francis for medium head (10–350 m), Kaplan/propeller for low head (2–40 m). Crossflow (Michell–Banki) and Turgo are used in small plants.
  2. Specific speed: Ns=NP/H5/4N_s = N\sqrt{P}/H^{5/4}; Pelton 10–35, Francis 60–300, Kaplan 300–1000 (metric, kW).
  3. Discharge and its variation: if flow varies greatly, choose a turbine with good part-load efficiency (Pelton, Kaplan) or use several units.
  4. Part-load operation: Francis efficiency falls below about 40–50% load; Pelton remains efficient down to 20%.
  5. Rotational speed: higher speed gives smaller generator; limited by cavitation and runaway speed.
  6. Cavitation: Thoma's coefficient fixes the setting below tailwater for reaction turbines.
  7. Sediment and abrasion: important for Himalayan rivers.
  8. Cost, efficiency, maintenance and local experience.

Example

For a site with Hnet=250H_{net} = 250 m and Q45=6Q_{45} = 6 m³/s split into two units of 3 m³/s, unit power ≈9.81×3×250×0.9≈6.6\approx 9.81 \times 3 \times 250 \times 0.9 \approx 6.6 MW. At 750 rpm, Ns=7506622/2505/4≈61N_s = 750\sqrt{6622}/250^{5/4} \approx 61, at the boundary of multi-jet Pelton and slow Francis; the Pelton is preferred if the river is silty or long part-load running is expected.

  • 2072 Asoj · 8 marks

Discuss the effect of the following factors in selecting of a turbine for a hydroelectric plant? (i) head (ii) speed and specific speed (iii) part load operation.

Answer

Turbine selection depends mainly on head, speed/specific speed and the expected part-load operation.

(i) Head

Head is the first and most important factor because it decides the velocity of water.

Head rangeSuitable turbine
High (above ~300 m, up to 1300 m)Pelton (impulse)
Medium (30–300 m)Francis (reaction)
Low (2–40 m)Kaplan / propeller
Small plants, 3–250 mCrossflow, Turgo
  • High head gives high jet velocity; an impulse turbine with small flow suits it.
  • Low head needs large flow to give the same power, so a large flow-area axial runner (Kaplan) is used.
  • Head variation also matters: Kaplan copes with varying head better than propeller.

(ii) Speed and specific speed

Ns=NPH5/4N_s = \frac{N\sqrt{P}}{H^{5/4}}
  • Specific speed joins head, power and speed into one value that fixes the runner shape: Pelton 10–35, Francis 60–300, Kaplan 300–1000.
  • A higher running speed gives a smaller, cheaper generator (fewer poles, N=120f/pN = 120f/p), so designers prefer the highest speed that is safe.
  • But higher NsN_s means a higher Thoma cavitation coefficient, so the runner must be set lower (more excavation), and the runaway speed is higher.
  • Therefore the speed is chosen as the highest synchronous speed for which cavitation and runaway limits are satisfied.

(iii) Part-load operation

ROR plants often run at part load in the dry season, so the shape of the efficiency curve matters.

 eff
  |   ____________ Pelton/Kaplan (flat)
  |  /    ___
  | /    /   \__ Francis (peaked)
  |/    /       \__ Propeller (sharp)
  +--------------------> % load
  0   25  50  75  100
  • Pelton: efficiency stays high from about 20% to 100% load (flow controlled by needle; multiple jets can be shut).
  • Kaplan: adjustable runner blades and guide vanes keep high efficiency over a wide range.
  • Francis: best efficiency near 80–90% of rated load; falls quickly below about 40–50%, with vibration and cavitation.
  • Propeller (fixed blade): very poor at part load.

If part-load running is expected, choose Pelton/Kaplan, or use several smaller Francis units so each runs near full load.

  • 2079 Jestha · 8 marks

Describe the characteristics of various types of turbines used in hydro electric power stations.

Answer

Hydraulic turbines are of two main kinds: impulse turbines, in which the whole pressure head is turned into a free jet before it strikes the runner (Pelton, Turgo, crossflow), and reaction turbines, in which the runner is full of water under pressure and both pressure and kinetic energy act on it (Francis, Kaplan, propeller).

Pelton turbine (impulse, tangential flow)

  • Head: high, about 50–1300 m; small discharge.
  • Specific speed: about 10–35 per jet (up to ~60 with multiple jets).
  • Water from one or more nozzles strikes double-cup buckets at atmospheric pressure; flow controlled by a spear (needle); a deflector cuts the jet on load rejection.
  • Flat efficiency curve: high efficiency from about 20% to 100% load. Maximum efficiency about 90–92%.
  • No draft tube; no cavitation in the runner; easy to inspect and repair, tolerant of sediment.

Francis turbine (reaction, mixed flow)

  • Head: medium, about 10–350 m (most common 30–300 m); medium discharge.
  • Specific speed: about 60–300.
  • Spiral casing, stay vanes, adjustable guide vanes, runner (radial inflow, axial outflow) and draft tube.
  • Highest peak efficiency (about 93–95%) but falls at part load below about 40–50%; vibration and draft tube surges at low load.
  • Sensitive to cavitation and sediment erosion; runner setting fixed by Thoma's σ\sigma.

Kaplan turbine (reaction, axial flow)

  • Head: low, about 2–40 m; large discharge.
  • Specific speed: about 300–1000.
  • Adjustable runner blades and guide vanes (double regulation) keep efficiency high over a wide load and head range.
  • Large size, high speed relative to head; high cavitation risk, so deep setting.
  • Propeller turbine: same with fixed blades; cheap but poor part-load efficiency.

Small-hydro turbines

  • Crossflow (Michell–Banki): head 3–250 m, simple, cheap, flat efficiency (about 75–85%); used in Nepalese micro-hydro.
  • Turgo: impulse, head 50–250 m, higher speed than Pelton for the same head.

Summary

FeaturePeltonFrancisKaplan
TypeImpulseReactionReaction
Flow directionTangentialRadial in, axial outAxial
HeadHighMediumLow
DischargeLowMediumHigh
NsN_s (kW)10–3560–300300–1000
Part-load efficiencyVery goodPoorVery good
Draft tubeNoYesYes
Cavitation riskLowMediumHigh
  • 2078 Chaitra · 4 marks

Differentiate applicability of Francis and Kaplan Turbine in reference to hydropower project.

Answer

Both Francis and Kaplan are reaction turbines, but the Francis is a mixed-flow turbine for medium heads, while the Kaplan is an axial-flow turbine with adjustable blades for low heads and large discharge.

PointFrancisKaplan
FlowRadial in, axial out (mixed)Axial
Head rangeAbout 10–350 m (typically 30–300 m)About 2–40 m
DischargeMediumLarge
Specific speed (kW)60–300300–1000
Runner bladesFixed, 9–19 vanesAdjustable, 3–8 blades
RegulationGuide vanes onlyGuide vanes and runner blades (double)
Part-load efficiencyFalls below ~40–50% loadHigh over wide range
CavitationModerateHigher; needs deep setting
Typical projectMedium-head ROR/PROR in hillsBarrage or canal-drop plant on large rivers

In hydropower projects: Francis suits most medium-head schemes in Nepal's hills (e.g. Kaligandaki "A", Middle Marsyangdi). Kaplan suits low-head, high-flow sites such as barrages on large rivers in the plains, canal falls and run-of-river schemes with large and varying flow.

  • 2080 Chaitra · 4 marks

Discuss about suitability of Kaplan turbine in context of Nepalese hydropower sector.

Answer

The Kaplan turbine suits low heads (about 2–40 m) with large discharge. Its suitability in Nepal is limited, because most Nepalese hydropower sites are in steep hills with medium to high heads.

Points against wide use

  • Nepal's rivers fall steeply, so most projects have heads of 50–600 m, which suit Francis and Pelton turbines.
  • Low-head sites lie mainly in the Terai and inner valleys, where the land is flat; a barrage there floods farmland and may affect India downstream (border rivers).
  • Nepalese rivers carry heavy sediment in the monsoon; Kaplan runners have high blade tip speeds and are prone to abrasion and cavitation.
  • Kaplan units are large and heavy for their output; transporting them on hill roads is hard, and civil works (deep setting) are costly.

Where Kaplan is suitable

  • Canal-drop and barrage plants: e.g. the Gandak hydropower station on the Gandak canal and potential sites on the Koshi, Karnali and Narayani in the plains.
  • Low-head, high-flow ROR schemes on large rivers in the Terai and Chure foothills.
  • Kaplan's flat part-load efficiency is useful where river flow varies a lot between seasons.
  • Small low-head schemes can use simpler propeller or bulb turbines.

So Kaplan has a niche role in Nepal; Francis and Pelton dominate, but Kaplan is the right choice for low-head, high-discharge sites in the southern plains.

  • 2073 Magh · 8 marks

What are the disadvantages of a very low specific speed reaction turbine? What are its advantages? How does the efficiency of the pelton wheel vary with its speed?

Answer

A very low specific speed reaction turbine is a slow Francis runner (about NsN_s = 60–100, kW units) used near the top of the Francis head range. Its runner is large in diameter with long, narrow passages and nearly radial flow.

Disadvantages

  1. Large runner diameter for the power produced, so the turbine and generator (more poles at low speed) are heavy and costly.
  2. High friction and disc-friction losses in long narrow passages, so peak efficiency is lower than a medium-NsN_s Francis.
  3. Low running speed means a large generator with many poles.
  4. Leakage losses through seals rise at high head.
  5. At heads where low-NsN_s Francis is used, a Pelton may be cheaper and more efficient at part load.

Advantages

  1. Low cavitation risk: Thoma's critical σ\sigma is small at low NsN_s, so the runner can be set higher above tailwater with less excavation.
  2. Can work under high heads (up to about 350–600 m), with small discharge.
  3. Robust runner, stable operation and smaller draft tube.
  4. Less sensitive to tailwater level changes.

Efficiency of Pelton wheel with speed

For a Pelton wheel with jet velocity V1V_1, bucket speed uu, blade outlet angle ϕ\phi and friction factor kk, the hydraulic efficiency is

ηh=2u (V1−u)(1+kcos⁡ϕ)V12\eta_h = \frac{2u\,(V_1 - u)(1 + k\cos\phi)}{V_1^2}

Setting dηhdu=0\dfrac{d\eta_h}{du} = 0:

V1−2u=0u=V12ηh,max=1+kcos⁡ϕ2\begin{aligned} V_1 - 2u &= 0 \\ u &= \frac{V_1}{2} \\ \eta_{h,max} &= \frac{1 + k\cos\phi}{2} \end{aligned}
  • Efficiency is a parabola in speed ratio u/V1u/V_1: zero when the wheel is stationary (u=0u=0), maximum at u/V1=0.5u/V_1 = 0.5 (in practice about 0.46 due to losses), and zero again at runaway (u=V1u = V_1).
  • With k=1k = 1 and ϕ=15∘\phi = 15^\circ, ηh,max=(1+cos⁡15∘)/2≈0.983\eta_{h,max} = (1+\cos 15^\circ)/2 \approx 0.983 (theoretical).
 eff
  |      ___
  |    /     \
  |   /       \
  |  /         \
  | /           \
  +--------------+---> u/V1
  0     0.46    1.0

Since a Pelton runs at constant synchronous speed while the jet velocity is fixed by head, the speed ratio stays near optimum, and load changes are handled by the needle; this is why its efficiency is high over a wide load range.

  • 2078 Kartik · 6 marks

What are hydrograph and flow duration curve? Discuss its applicability.

Answer

Hydrograph

A hydrograph is a graph of river discharge plotted against time in calendar order (hours, days or months). It shows when high and low flows occur, e.g. the monsoon peak in July–August and the dry-season low in March–April for Nepalese rivers.

 Q (m3/s)
  |            __
  |           /  \
  |          /    \
  |         /      \__
  |  _____/            \______
  +---------------------------> Month
   Jan  Mar  May  Jul  Sep  Nov

Flow duration curve (FDC)

An FDC is a graph of discharge against the percentage of time it is equalled or exceeded. It is made by sorting the same flow data in descending order and plotting each value against its exceedance P=mn+1×100%P = \dfrac{m}{n+1}\times 100\%. Time order is lost, but the duration of each flow is shown.

 Q (m3/s)
  |\
  | \
  |  \___
  |      \______
  |             \__________
  +--------------------------> % time exceeded
  0     45   65       95  100

Applicability of the hydrograph

  1. Seasonal energy: monthly flows give month-wise energy, needed for wet/dry season tariff and PPA.
  2. Flood design: peak flood hydrographs size spillways, cofferdams and diversion works.
  3. Storage design: the mass curve (cumulative hydrograph) gives the reservoir or pondage volume needed to meet demand.
  4. Construction and maintenance planning: plan works and unit outages in low-flow months.
  5. Sediment planning: high-flow months carry most sediment, guiding desander operation.

Applicability of the FDC

  1. Design discharge: chosen at a set exceedance, e.g. Q40Q_{40}–Q45Q_{45} for ROR in Nepal.
  2. Installed capacity and annual energy: area under the FDC up to the design flow gives energy; plant factor is found from it.
  3. Firm power: from Q90Q_{90}–Q95Q_{95}.
  4. Number and size of units: shows how often the plant runs at part load.
  5. Riparian release: from the low-flow tail.
  6. Comparison of sites: a flat FDC means steady flow, ideal for ROR; a steep FDC means storage is needed.
PointHydrographFDC
x-axisTime (calendar)% time exceeded
Shows sequenceYesNo
Main useSeasonal energy, floods, storageDesign flow, capacity, firm power
  • 2078 Chaitra · 4 marks

How would flow duration curve affect while designing unit sizes of power plant?

Answer

The flow duration curve (FDC) shows how long each flow is available, so it decides how many units are needed and how big each should be, so that turbines run near their best efficiency for most of the year.

How the FDC affects unit sizing

  1. Total design flow: the plant's design discharge is taken at a chosen exceedance (e.g. Q45Q_{45}). This fixes total capacity P=ηρgQHP = \eta \rho g Q H.
  2. Minimum turbine flow: every turbine has a minimum stable flow (Francis about 40% of rated, Pelton about 10–20%, Kaplan about 20–30%). The dry-season flow from the FDC (e.g. Q85Q_{85}–Q95Q_{95}) must be at least the minimum flow of one unit, or the plant must shut down in the dry season.
  3. Steep FDC → more units: if the flow drops far below the design flow for long periods, divide the capacity into two or more units. In the dry season one unit runs near full load instead of all units at poor part load.
  4. Flat FDC → fewer, larger units: steady flow allows one or two large units, which are cheaper per kW.
  5. Equal-sized units are preferred for interchangeable spares and simple operation; unequal units are used only when the FDC is very steep.
  6. Reliability and maintenance: with at least two units, maintenance can be done in the low-flow season while the other unit uses all the available water.

Example

If Q45=10Q_{45} = 10 m³/s and Q95=2.2Q_{95} = 2.2 m³/s, a single Francis unit would need at least 0.4×10=40.4 \times 10 = 4 m³/s and would stop in the driest months. With two units of 5 m³/s each, one unit needs only 2 m³/s, so it keeps running even at Q95Q_{95}.

  • 2075 Bhadra · 6 marks

The design exceedance of hydropower plant Q45%, what does that mean? Compare Q45% with Q65% while designing hydropower plant from the perspective of power producer (seller).

Answer

Meaning of Q45%

A design exceedance of Q45% means the plant's design (rated) discharge is the flow that the river equals or exceeds 45% of the time, about 0.45×365≈1640.45 \times 365 \approx 164 days a year. The installed capacity is based on this flow:

P=ηoρg (Q45−Qriparian) HnetP = \eta_o \rho g\,(Q_{45} - Q_{riparian})\,H_{net}

For the other 55% of the time the plant runs at part load, following the river flow. In Nepal, NEA's power purchase practice for ROR projects has moved from about Q65Q_{65} (older practice) to about Q40Q_{40}–Q45Q_{45}.

Comparison from the seller's (IPP) point of view

PointQ45% designQ65% design
Design flowLargerSmaller
Installed capacityHigher (MW)Lower
Annual energyMore (extra wet-season energy)Less
Plant factorLower (about 55–65%)Higher (about 70–80%)
Capital costHigher (bigger waterway, units)Lower
Cost per kWhCan be higher if the extra energy is only wet-seasonLower per kWh
Wet-season spill/curtailment riskHigher (NEA may curtail surplus in monsoon)Lower
Dry-season share of energySmaller proportionLarger proportion
RevenueHigher total, if all energy is boughtLower total, steadier

Seller's view

  • Q45 gives more MW and more total energy, which raises revenue if NEA buys all energy under a take-or-pay PPA. Bank financing and per-MW benefits also favour larger capacity.
  • But most of the extra energy comes in the monsoon, when the tariff is lower (dry-season rate is higher than wet-season rate in NEA's posted rates) and when the grid already has surplus, so spillage or curtailment can reduce income.
  • Q65 gives a smaller, cheaper plant with higher plant factor and lower risk, but loses wet-season energy.
  • The final choice is made by comparing incremental energy and revenue against incremental cost (B/C ratio, IRR) for several exceedance values; with a firm PPA, sellers usually prefer about Q40–Q45.
  • 2074 Bhadra · 5 marks

Why riparian release is considered during the design of hydropower plant? What are the criteria of design discharge selection?

Answer

Why riparian release is considered

Riparian release (environmental flow, compensation flow) is the minimum flow that must be left in the river below the diversion weir. It is deducted from the river flow before calculating the plant's design discharge.

  • Aquatic life: keeps fish, insects and plants alive in the dewatered stretch between weir and tailrace; allows fish migration.
  • Downstream water rights: people use river water for drinking, irrigation, water mills, washing and cremation (religious use) at ghats.
  • Water quality and landscape: prevents stagnation, pollution and loss of scenic value.
  • Legal requirement: in Nepal, the Hydropower Development Policy 2001 and EIA guidelines require releasing at least 10% of the minimum monthly average flow, or more if the EIA finds it necessary.

So available flow for generation =Qriver−Qriparian= Q_{river} - Q_{riparian}.

Criteria for selection of design discharge

  1. Flow duration curve: design flow is chosen at a set exceedance, e.g. Q40Q_{40}–Q45Q_{45} for ROR in Nepal (older practice Q65Q_{65}).
  2. Riparian release deducted first.
  3. Type of project: ROR uses an exceedance flow; PROR and storage projects can use larger flows for peaking.
  4. Economics: design flow is increased step by step until the incremental benefit equals incremental cost (maximum NPV or acceptable IRR, B/C).
  5. PPA and grid demand: the buyer's policy (NEA) on exceedance, dry/wet energy and plant factor.
  6. Minimum turbine flow and number of units: the low-flow season must still run at least one unit.
  7. Sediment and site constraints: tunnel size, desander capacity and geology limit the flow that can be conveyed economically.
  • 2078 Chaitra · 4+2 marks

What are the factors to be considered during selection of appropriate site for hydropower? Also mentions the factors that govern discharge of a hydropower project.

Answer

Factors for selection of a hydropower site

  1. Availability of water: a river with good, reliable flow (from rainfall, snow and glaciers), shown by long-term hydrological data.
  2. Available head: a steep river reach or a river bend where a short tunnel gives a large head.
  3. Geology: sound rock for dam, tunnel and powerhouse; avoid active faults, landslides and weak zones.
  4. Storage possibility: a narrow gorge with a wide valley upstream if storage is wanted.
  5. Sediment load: low sediment or space for a desander.
  6. Access and transport: roads for heavy equipment and construction materials.
  7. Distance to load or grid: short transmission line reduces cost and losses.
  8. Environmental and social impact: minimal submergence, resettlement, forest loss and impact on protected areas.
  9. Natural hazards: floods, GLOF, earthquakes, debris flows.
  10. Cost of the project: cost per kW and per kWh.

Factors governing discharge of a hydropower project

  1. Catchment area above the intake.
  2. Rainfall amount and distribution (monsoon), plus snow and glacier melt.
  3. Evaporation and transpiration losses.
  4. Catchment characteristics: slope, soil, geology, forest cover, which control runoff and baseflow.
  5. Upstream use: irrigation and drinking water abstractions, and riparian release requirement.
  • 2070 Magh · 5+3 marks

What are the criteria for selection of a hydropower project? If the water in the river is known to carry high sediments and debris, what would be your particular recommendation in its design?

Answer

Criteria for selection of a hydropower project

  1. Hydrology: sufficient and reliable flow; FDC and long-term data.
  2. Head: steep gradient or large drop over a short distance.
  3. Topography and geology: good dam, tunnel and powerhouse sites; stable slopes.
  4. Sediment and hazards: sediment load, floods, GLOF, landslides and seismicity.
  5. Access and transmission: road access and distance to grid or load centre.
  6. Environmental and social impact: submergence, resettlement, fish, protected areas; EIA clearance.
  7. Power market: demand, PPA possibility and energy mix (peaking or base load).
  8. Economics: cost per kW, levelised cost per kWh, IRR, B/C ratio, payback.
  9. Legal and institutional: licences, water rights and local acceptance.

Recommendation for a river with high sediment and debris

  1. Headworks: gated weir with undersluice to keep the intake area flushed; boulder/debris barrier and coarse trash rack upstream of intake.
  2. Intake design: side intake placed on the outer bend, raised sill and sediment excluder to draw cleaner surface water.
  3. Gravel trap close to the intake with flushing gates.
  4. Larger desander: multiple bays, designed to remove finer particles (e.g. 0.2 mm, or 0.1–0.15 mm for high-head Pelton/Francis), with continuous or frequent flushing.
  5. Turbine choice: prefer Pelton where head allows, as it is easier to repair; for Francis, use lower runner velocity and sediment-resistant design.
  6. Materials: 13Cr-4Ni stainless steel runners with hard coatings (e.g. tungsten-carbide HVOF) on runners, guide vanes and nozzles.
  7. Operation: monitor sediment concentration online and shut down units during flood peaks above a set limit; keep spare runners and needles.
  8. For storage/pondage: provide low-level flushing outlets and plan regular reservoir flushing.

Questions from Old Question Collection (EE 753) (IOE EE 753 exam papers from 2070 Magh to 2082 Shrawan), Question bank (ioesolutions) (IOE EE 753 exam papers from 2070 Bhadra to 2074 Magh) and Old questions (NCE Library) (IOE EE 753 exam papers from 2070 Bhadra to 2080 Chaitra). Answers are written for this site; check them against your class notes.

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