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Chapter 4

Hydrology and Hydropower

Lok Sewa Aayog (Public Service Commission) past exam questions

Past questions and answers

7 questions set from this chapter. Most repeated first.

  • 2083 · 1st · 5 marks

Discuss the key prospects and challenges for hydropower development in Nepal. 5

Answer

Key prospects for hydropower development in Nepal

  1. Huge technically feasible potential — about 83,000 MW technically and 43,000 MW economically feasible, against an installed capacity of only around 3,000 MW, so the resource is far from exhausted.
  2. High specific run-off and steep gradient — steep north–south flowing rivers over short distances give high head, lowering cost per MW and favouring run-of-river as well as storage schemes.
  3. Growing domestic demand and energy security — rising industrial, commercial and household demand, plus substitution of imported fossil fuel and reduction of the trade deficit through electricity export.
  4. Regional export market — surplus seasonal energy can be sold to India and, through it, to Bangladesh under bilateral power trade agreements, earning foreign exchange.
  5. Clean energy and climate commitments — hydropower supports NDC targets, carbon credit earning and climate finance (Green Climate Fund, carbon markets).
  6. Policy and institutional support — Electricity Act, 2049; Electricity Regulatory Commission; one-door clearance; FIT/PPA arrangements; and donor/MDB financing.

Key challenges

  1. Financial constraints — large capital requirement, weak domestic capital market, high cost of debt, difficulty in financial closure and in raising equity for storage projects.
  2. Hydrological and climate risk — seasonal flow variation (wet–dry disparity), glacial lake outburst floods (GLOF), landslides and sedimentation reducing reservoir life.
  3. Infrastructure deficit — inadequate transmission lines and cross-border evacuation capacity, poor road access to remote sites, which delays commissioning and raises cost.
  4. Social and resettlement issues — land acquisition, compensation disputes, local demands for shares/royalty/employment, and delays due to local obstruction.
  5. Policy instability and procedural delays — frequent changes in policy, licence cancellation and re-issue, slow environmental clearance and PPA approval.
  6. Technical capacity gap — shortage of experienced manpower in design, construction and operation of large and storage projects; dependence on foreign contractors.
  7. Market and pricing risk — uncertain export demand and pricing, take-or-pay issues with India, and weak domestic purchasing power to absorb tariff.

Conclusion — Nepal's hydropower prospects are strong, but realising them requires stable policy, transmission investment, transparent local benefit-sharing and access to affordable long-term finance.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 5 marks

Explain firm and secondary energy with regard to load curve in Nepal.

Answer

Load curve: a plot of demand (MW) against time (daily, monthly, annual); the area under it is the energy consumed. Nepal's daily curve has a morning peak and a larger evening peak (about 6-8 PM, with lighting and domestic load), and a deep valley at night.

Firm energy is the energy that a hydropower plant or system can deliver reliably (about 95-100% of the time, or in the driest year) in the critical dry season (Poush-Baisakh, roughly December to May), when river flow is lowest. It is the guaranteed energy on which supply contracts (PPAs) and planning are based. On the load curve it corresponds to the part of the demand that must be served continuously, mainly the base-load band, and in Nepal it comes from run-of-river plants at their dry-season flows and from storage plants such as Kulekhani.

Secondary energy is the energy a plant can generate in excess of its firm energy, when flow exceeds the firm design flow, mainly in the monsoon (June-November). It is not guaranteed and is sold at a lower tariff. It can serve load above the firm level, interruptible and seasonal industrial loads, or export to India; otherwise it is spilled. (It is a supply-side surplus, not simply the area of the demand curve above the base-load band.)

In Nepal:

  • Most plants are run-of-river, so firm energy is small compared with installed capacity: surplus (secondary) energy in the wet season but shortage and imports from India in the dry season.
  • Storage reservoir projects (e.g. Tanahu, Budhi Gandaki) raise firm energy by shifting wet-season flow into the dry season, converting secondary energy into firm energy and allowing peaking operation.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Identify all losses in the hydraulic system of a typical hydropower project involving tunnel system for water conveyance. Discuss steps involved in calculation of Net-Head for hydropower project.

Answer

Hydraulic Losses in the Water Conveyance Tunnel System of a Hydropower Project

1. Losses in the Hydraulic System

In a typical hydropower project, water is conveyed from the intake to the powerhouse through a tunnel system (headrace tunnel, surge shaft, penstock, etc.). The total head available at the intake is progressively reduced by the following losses:

A. Major (Friction) Losses

  1. Friction loss in the headrace tunnel – Loss due to friction between flowing water and the tunnel wall, given by the Darcy–Weisbach equation:

hf1=f1L1D1V122gh_{f1} = f_1 \frac{L_1}{D_1} \frac{V_1^2}{2g}

where f1f_1 = friction factor, L1L_1 = length, D1D_1 = diameter (or hydraulic diameter for non-circular tunnels), V1V_1 = flow velocity.

  1. Friction loss in the penstock – Similarly,

hf2=f2L2D2V222gh_{f2} = f_2 \frac{L_2}{D_2} \frac{V_2^2}{2g}

  1. Friction loss in the surge shaft / surge tank – Minor but present, especially during transient flow.

B. Minor (Local) Losses

These occur at changes in geometry or flow direction and are expressed as:

hm=KV22gh_m = K \frac{V^2}{2g}

where KK is the loss coefficient. They include:

  1. Entrance loss at the intake (bell-mouth entry): K≈0.05K \approx 0.05–0.50.5 depending on entry shape.
  2. Trash rack loss at the intake: htr=KtrV22gh_{tr} = K_{tr} \frac{V^2}{2g}, with KtrK_{tr} depending on bar spacing, bar thickness, and clogging.
  3. Bend losses in the tunnel/penstock due to curvature: KK depends on deflection angle and radius of curvature.
  4. Contraction and expansion losses where the tunnel cross-section changes.
  5. Gate and valve losses (e.g., at the inlet valve, butterfly valve).
  6. Junction/bifurcation losses at the surge shaft, bifurcation, or manifold.
  7. Draft tube exit loss – The velocity head at the draft tube outlet discharged to the tailrace, of which only a portion is recovered; the unrecovered part constitutes a loss.
  8. Tailrace losses – Friction and minor losses in the tailrace channel/tunnel downstream of the turbine.

C. Other Losses

  1. Loss due to roughness and aging of the tunnel lining (increased ff over time).
  2. Loss due to sediment abrasion of the tunnel/penstock surface.
  3. Loss due to leakage through gates, seals, and joints.
  4. Loss due to air entrainment and flow separation at poorly designed transitions.

The total head loss is the sum:

∑hL=hf1+hf2+∑hm\sum h_L = h_{f1} + h_{f2} + \sum h_m


2. Steps Involved in Calculation of Net Head

Net head (HnetH_{net}) is the effective head available at the turbine after deducting all losses from the gross head.

Step 1: Determine the Gross Head (HgrossH_{gross})

Hgross=Zintake−ZtailraceH_{gross} = Z_{intake} - Z_{tailrace}

where ZintakeZ_{intake} = water level at the intake (forebay), and ZtailraceZ_{tailrace} = water level at the tailrace outlet. For a run-of-river plant, the design (minimum) flow level is used.

Step 2: Identify and Compute Major (Friction) Losses

Using the Darcy–Weisbach equation for each conduit (headrace tunnel, penstock, etc.):

hf=fLDV22gh_f = f \frac{L}{D} \frac{V^2}{2g}

  • Determine ff from the Moody diagram or Colebrook–White equation using Reynolds number and relative roughness.
  • Compute velocity V=Q/AV = Q/A for the design discharge QQ.
  • For non-circular tunnels, use the hydraulic diameter Dh=4A/PD_h = 4A/P.

Step 3: Compute Minor (Local) Losses

For each fitting, bend, transition, trash rack, valve, etc.:

hm=KV22gh_m = K \frac{V^2}{2g}

Sum all minor losses: ∑hm\sum h_m.

Step 4: Compute Total Head Loss

∑hL=∑hf+∑hm\sum h_L = \sum h_f + \sum h_m

Step 5: Calculate Net Head

Hnet=Hgross−∑hLH_{net} = H_{gross} - \sum h_L

Step 6: Apply Correction for Tailrace and Other Effects (if required)

  • Deduct tailrace losses.
  • Account for the unrecovered velocity head at the draft tube exit if not already included.
  • For reaction turbines, the net head is measured from the forebay water level to the tailwater level minus losses.

Step 7: Verify and Finalize

  • Check that HnetH_{net} is positive and sufficient for the selected turbine type.
  • Compare with the design head used for turbine selection.
  • If HnetH_{net} is too low, redesign the conduit (increase diameter, reduce length, or smooth lining).

Final Expression:

Hnet=(Zintake−Ztailrace)−[fLDV22g+∑KV22g]\boxed{H_{net} = (Z_{intake} - Z_{tailrace}) - \left[ f \frac{L}{D} \frac{V^2}{2g} + \sum K \frac{V^2}{2g} \right]}

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

What are the different types of Turbines used in hydropower projects? What are the key factors that govern selection of appropriate turbine? Also, discuss the types of generators and their use in brief.

Answer

Types of Turbines Used in Hydropower Projects

Turbines are classified mainly on the basis of the action of water on the runner (impulse vs. reaction) and the direction of flow.

A. Impulse Turbines (pressure remains constant; energy converted by velocity)

  1. Pelton Wheel – Water strikes bucket-shaped vanes through one or more jets. Used for high head (> 250 m) and low discharge. Most common in Nepal's high-head projects (e.g., Khimti, Chilime).
  2. Turgo Wheel – A modified Pelton with a jet striking the runner at an angle; handles medium-to-high head with larger flows than Pelton.
  3. Cross-flow (Banki–Michell) Turbine – Water passes twice through the runner blades. Simple, cheap, suited to low head (5–100 m) and small micro-hydro plants.

B. Reaction Turbines (pressure drops across the runner; both pressure and velocity energy used)

  1. Francis Turbine – Mixed-flow (radial inlet, axial outlet). Used for medium head (10–350 m) and medium-to-large discharge. Widely used in Nepal (e.g., Kulekhani, Middle Marsyangdi).
  2. Propeller / Kaplan Turbine – Axial flow. Propeller has fixed blades for low head (2–20 m) with nearly constant flow; Kaplan has adjustable blades for low head with widely varying flow (e.g., Gandak, Koshi barrage plants).
  3. Bulb / Tubular Turbine – Generator housed in a bulb in the water passage; used for very low head (1–20 m) in run-of-river and tidal schemes.

Key Factors Governing Selection of Appropriate Turbine

  1. Head (net head) – The single most important factor. High head → Pelton; medium head → Francis; low head → Kaplan/Propeller.
  2. Discharge (flow rate) and its variation – Determines runner size and whether adjustable blades (Kaplan) are needed for part-load efficiency.
  3. Specific speed (Ns) – Correlates head and discharge; each turbine type has a characteristic Ns range (Pelton low, Francis medium, Kaplan high).
  4. Power capacity and load factor – Base-load vs. peak-load operation affects choice of adjustable-blade machines.
  5. Efficiency over the operating range – Must remain high at part load, not only at design point.
  6. Cavitation and setting (suction head) – Reaction turbines need proper submergence; NPSH must be satisfied.
  7. Cost, size and civil works – Pelton needs smaller powerhouse; Kaplan needs larger, deeper foundations.
  8. Sediment content and water quality – High sediment favours Pelton (abrasion-resistant buckets) over Francis.
  9. Speed and generator matching – Turbine speed should match synchronous generator speed or gear ratio.
  10. Availability, maintenance and manufacturer's standard sizes.

Types of Generators and Their Use

  1. Synchronous Generator (Alternator) – Most widely used in hydropower. Runs at synchronous speed (N = 120f/P) and supplies both active and reactive power; can control voltage and power factor. Used in all grid-connected plants. Types: salient-pole (low speed, directly coupled to Kaplan/Pelton) and cylindrical rotor (high speed, used with steam/gas turbines).
  2. Induction (Asynchronous) Generator – Simple, cheap, rugged; needs no separate excitation and synchronising. Used in micro/mini-hydro and small isolated plants; cannot supply reactive power (draws it from the grid).
  3. Permanent Magnet Synchronous Generator (PMSG) – Compact, high efficiency, no slip rings; increasingly used in micro-hydro and variable-speed applications.
  4. DC Generator – Rare in modern hydropower; used only in very small isolated plants or for excitation (exciter) purposes.

In brief: For grid-connected hydropower, the synchronous alternator is standard; for isolated small plants, induction or PMSG units are preferred for their simplicity and low cost.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

What is understood by the optimization of any components of hydropower projects? What are the objectives of optimization? Explain with mathematical expression the optimization of Penstock.

Answer

Optimization of Hydropower Components

Optimization in hydropower engineering means selecting the size, type and layout of a component (penstock, tunnel, turbine, canal, etc.) so that the total cost over the life of the project is minimum while satisfying the required discharge, head and safety conditions. Every component has two opposing cost tendencies:

  • If the component is made larger (bigger diameter, wider canal), the capital cost increases but the energy loss (head loss) decreases, so more energy is sold every year.
  • If the component is made smaller, the capital cost decreases but the head loss increases, so annual revenue from energy decreases.

The optimum is the size at which the sum of capital cost (annualised) and the cost of lost energy is minimum.

Objectives of Optimization

  1. To obtain the minimum total annual cost (capital + operation + energy loss cost) of the component and hence of the project.
  2. To maximize the net benefit / internal rate of return of the project.
  3. To select the most economical diameter, cross-section, material and alignment of waterways.
  4. To reduce head losses so that the firm energy and peak energy output is maximized.
  5. To make the best use of available head, discharge and topography within structural and safety limits.
  6. To arrive at a design that is technically feasible and economically justified (benefit–cost ratio > 1).

Mathematical Optimization of Penstock

Let,

SymbolQuantityUnit
QQdesign dischargem³/s
HHgross headm
DDinternal diameter of penstockm
LLlength of penstockm
ffDarcy–Weisbach friction factor–
hfh_ffriction head lossm
nnnumber of annual operating hoursh
CeC_etariff / selling price of energyRs per kWh
CsC_scost of steel per tonneRs
ttwall thickness of penstockm
ρs\rho_sdensity of steelkg/m³
ρ\rhodensity of waterkg/m³
rrannual discount rate–
NNlife of plantyears

1. Head loss in the penstock (Darcy–Weisbach):

hf=f L v22gD,v=4QπD2h_f = \frac{f\,L\,v^{2}}{2gD}, \qquad v = \frac{4Q}{\pi D^{2}}

Substituting vv:

hf=8 f L Q2π2 g D5=KD5,K=8fLQ2π2gh_f = \frac{8\,f\,L\,Q^{2}}{\pi^{2}\,g\,D^{5}} = \frac{K}{D^{5}}, \qquad K = \frac{8fLQ^{2}}{\pi^{2}g}

2. Annual energy lost due to this head loss:

Elost=γ Q hf1000×n(kWh/year)E_{lost} = \frac{\gamma\,Q\,h_f}{1000}\times n \quad \text{(kWh/year)}

where γ=ρg=9810\gamma = \rho g = 9810 N/m³. Hence the annual cost of lost energy:

Cenergy=Ce⋅Elost=Ce⋅γ Q n1000⋅KD5C_{energy} = C_e \cdot E_{lost} = C_e \cdot \frac{\gamma\,Q\,n}{1000}\cdot \frac{K}{D^{5}}

3. Capital cost of the penstock. The weight of steel is

W=πD t L ρsW = \pi D \, t \, L \, \rho_s

With thickness from the hoop-stress formula t=pD2σallow+tcorrt = \dfrac{pD}{2\sigma_{allow}} + t_{corr}, and p=ρgHmaxp = \rho g H_{max}, the weight (and hence cost) is approximately proportional to D2D^{2}:

Ccap=Cs⋅W=Cs⋅πDLρs(ρgHmaxD2σallow+tcorr)≈aD2+bDC_{cap} = C_s \cdot W = C_s \cdot \pi D L \rho_s \left(\frac{\rho g H_{max} D}{2\sigma_{allow}} + t_{corr}\right) \approx aD^{2} + bD

4. Annualised capital cost (using the capital recovery factor):

Cannual=Ccap×r(1+r)N(1+r)N−1=(aD2+bD)⋅CRFC_{annual} = C_{cap}\times \frac{r(1+r)^{N}}{(1+r)^{N}-1} = (aD^{2}+bD)\cdot CRF

5. Total annual cost to be minimized:

Ctotal(D)=(aD2+bD)⋅CRF⏟annualised capital cost+Ce⋅γQn1000⋅KD5⏟cost of lost energyC_{total}(D) = \underbrace{(aD^{2}+bD)\cdot CRF}_{\text{annualised capital cost}} + \underbrace{C_e \cdot \frac{\gamma Q n}{1000}\cdot \frac{K}{D^{5}}}_{\text{cost of lost energy}}

Condition for optimum diameter:

dCtotaldD=0\frac{dC_{total}}{dD} = 0

CRF (2aD+b)−5 Ce γQn1000⋅KD6=0CRF\,(2aD + b) - 5\,C_e\,\frac{\gamma Q n}{1000}\cdot\frac{K}{D^{6}} = 0

Neglecting the small linear term bDbD (i.e. taking b≈0b \approx 0):

2a D CRF=5 Ce γ Q n K1000 D62a\,D\,CRF = \frac{5\,C_e\,\gamma\,Q\,n\,K}{1000\,D^{6}}

D7=5 Ce γ Q n K2000 a CRFD^{7} = \frac{5\,C_e\,\gamma\,Q\,n\,K}{2000\,a\,CRF}

Substituting γ=ρg\gamma = \rho g and K=8fLQ2π2gK = \dfrac{8fLQ^{2}}{\pi^{2}g}:

D7=5 Ce ρg Q n2000 a CRF⋅8fLQ2π2g=40 Ce ρ f L Q3 n2000 π2 a CRF=20 Ce ρ f L Q3 n1000 π2 a CRFD^{7} = \frac{5\,C_e\,\rho g\,Q\,n}{2000\,a\,CRF}\cdot\frac{8fLQ^{2}}{\pi^{2}g} = \frac{40\,C_e\,\rho\,f\,L\,Q^{3}\,n}{2000\,\pi^{2}\,a\,CRF} = \frac{20\,C_e\,\rho\,f\,L\,Q^{3}\,n}{1000\,\pi^{2}\,a\,CRF}

Dopt=[20 Ce ρ f L Q3 n1000 π2 a CRF]1/7\boxed{D_{opt} = \left[\frac{20\,C_e\,\rho\,f\,L\,Q^{3}\,n}{1000\,\pi^{2}\,a\,CRF}\right]^{1/7}}

The second derivative at this value is positive, confirming a minimum. Thus the most economical penstock diameter varies as Q3/7Q^{3/7} (and as the seventh root of length and energy tariff), and inversely as the seventh root of the steel cost and discount rate. In practice DoptD_{opt} is checked against the limiting velocity (usually 2–5 m/s) and the head loss limit (generally 2–10 % of gross head).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

As per the "Guidelines for Study of Hydropower Projects" issued by DoED (Government of Nepal), what are the major study requirements related to topographical and construction material surveys for a medium sized (10-50MW) project? Explain.

Answer

DoED Guidelines: Topographical and Construction Material Surveys (10-50 MW Project)

The Guidelines for Study of Hydropower Projects of the Department of Electricity Development (DoED) require a medium project (10-50 MW) to be studied to feasibility level, with the surveys below as the basis of layout, quantities and cost estimate. The scales and intervals are those normally adopted under the guidelines.

A. Topographical Survey

Objectives: to give accurate maps and sections for fixing the layout of the headworks, waterway and powerhouse, and for estimating quantities.

AreaTypical map scaleContour interval
Whole project area (catchment, access, transmission corridor)1:25,000 - 1:50,000 (from Survey Department maps)10-20 m
Headworks (weir, intake, gravel trap, desander)1:500 - 1:1,0001-2 m
Headrace canal/tunnel alignment, surge tank, penstock1:1,000 - 1:2,000 (strip about 200-300 m wide)2-5 m
Powerhouse, switchyard and tailrace1:500 - 1:1,0001-2 m
Quarry, borrow and muck disposal areas, camp sites1:1,000 - 1:2,0002-5 m

Main requirements:

  1. Control survey: permanent benchmarks and control points tied to the national datum and grid (Survey Department network, GPS/total station), with traverse closure within permissible limits.
  2. Detailed topographic survey by total station/DGPS (supported by LiDAR or drone where suitable) of all main structure sites.
  3. River surveys: longitudinal profile of the river along the project reach and cross-sections at the headworks, at regular intervals (about 50-100 m) and at the tailrace, with high flood marks and rating-curve gauge levels.
  4. Longitudinal profiles and cross-sections of the waterway alignment (canal, tunnel, penstock) at fixed intervals (about 20-50 m) and at every change of slope or direction.
  5. Access road and transmission line route survey to the extent needed for the feasibility cost estimate.
  6. Preparation of layout drawings, plans and sections, with all survey data and maps submitted as a report annex.

B. Construction Material Survey

Objectives: to locate enough suitable material close to the site, so as to cut haul cost and secure quality.

  1. Identification of sources: quarries for rock aggregate and rip-rap, borrow areas for earth fill, impervious material (core) and sand/gravel from the river, with map location, access and haul distance.
  2. Investigation: test pits, trenches or auger holes and, where needed, drilling at each source, with sampling.
  3. Quantity estimation: proven and probable volumes, which should be about 1.5-2 times the requirement, allowing for waste and spoil.
  4. Laboratory testing per Nepal Standards, IS or ASTM:
    • Aggregate: gradation, specific gravity, water absorption, Los Angeles abrasion, soundness, alkali-aggregate reactivity.
    • Sand: fineness modulus, silt content.
    • Earth fill: grain size, Atterberg limits, compaction (Proctor), permeability and shear strength.
    • Rock: compressive strength.
  5. Suitability and economics: compare sources on quality, quantity, haul distance, environmental impact and royalty or permit conditions, then adopt a material sourcing plan.
  6. Muck disposal and cement/steel: identify spoil disposal sites for the excavated material, and the sources of cement, reinforcement and other imported items.

Conclusion

The topographical survey fixes the layout and quantities, and the construction material survey confirms that quality material is available at a reasonable cost. Together they give a reliable base for the design, construction planning and cost estimate of the feasibility report.

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from IDM Exam Question for Engineer 7th (Engineer 7th (Technical/Electrical) · 2nd · 2082), NEA Question for Engineer/Survey (Engineer/Survey-Engineer / Environmentalist 7th · 1st · 2083), NEA Question for Engineer 7th Lok Sewa (Engineer 7th Level · 1st · 2083), NEA Exam Question for Civil Engineer (Engineer 7th Level, Civil · 2nd · 2082) and NEA Lok Sewa Engineer/Survey Engineer (Engineer/ Survey Engineer (7th Level) · 1st · 2081). Answers are written for this site; check them against your class notes.