Planning of Hydropower Projects

A hydropower project is studied in stages of increasing detail, so that money is spent on detailed work only for projects that prove worthwhile. Together with construction and operation these form the hydropower development cycle.

 Identification / desk study
            |
 Reconnaissance study
            |
 Pre-feasibility study
            |
 Feasibility study  (+ EIA, licences)
            |
 Detailed design & tender
            |
 Construction & commissioning
            |
 Operation & maintenance  --> rehabilitation

1. Desk study / identification

Using maps (1:50 000 topo), existing hydrological records and reports, possible sites are identified and rough power potential $P = 9.81\,QH\eta$ is computed. Several alternatives are listed.

2. Reconnaissance study

A short site visit and study to screen alternatives.

• Rough layout: headworks, waterway, powerhouse locations.
• Approximate head (altimeter/GPS) and spot flow measurement.
• General geology, access, land use, settlements.
• Rough cost and power; ranking of sites.

3. Pre-feasibility study

Selects the best scheme and decides whether a full feasibility study is justified.

• Topographic survey at larger scale; hydrological analysis (flow duration curve, floods).
• Preliminary geology (surface mapping), sediment and seismic data.
• Comparison of layouts and capacities; preliminary design and cost estimate.
• Initial environmental examination, power market and economic indicators (B/C, IRR).

4. Feasibility study

Detailed enough for financing and licensing decisions.

• Detailed topographic surveys; long-term hydrology, design flood, sediment study.
• Subsurface investigation: drilling, test pits, seismic refraction, lab tests of construction materials.
• Optimisation of installed capacity, firm/secondary energy; final layout.
• Design of main structures, construction schedule, detailed cost estimate.
• Full EIA, social and resettlement study; power market and transmission; financial and economic analysis (NPV, IRR, B/C, cost per kWh).
• In Nepal, the generation licence (DoED) and PPA with NEA are based on this stage.

5. Detailed design and tender

Final engineering drawings, specifications, bill of quantities, tender documents; further investigation where needed.

6. Construction and commissioning

Access roads, river diversion, civil works, electro-mechanical and transmission works; testing and commissioning.

7. Operation and maintenance

Power generation, monitoring of sediment, structures and equipment; later upgrading or rehabilitation, which restarts the cycle.

The best (optimum) installed capacity is the capacity at which the net annual benefit (value of energy sold minus annual cost of the capacity) is maximum. Equivalently, the last increment of capacity added should earn at least as much as it costs.

Data and assumptions

• Each month is taken as $8760/12 = 730$ h.
• Energy price = 40 $/MWh = 0.04 $/kWh.
• Capital recovery factor (10%, 40 years):

$$\begin{aligned} CRF &= \frac{i(1+i)^n}{(1+i)^n-1} = \frac{0.10(1.10)^{40}}{(1.10)^{40}-1} = 0.10226 \end{aligned}$$

• Annual cost per kW of capacity = capital charge + O&M:

$$\begin{aligned} C_a &= 800\,(0.10226 + 0.025) = 101.81\ \text{\$/kW/yr} \end{aligned}$$

Power duration (descending order)

650, 560, 510, 400, 320, 270, 200, 140, 110, 95, 80, 65 kW. A capacity $P_c$ produces $\min(P_i, P_c)$ in each month.

Annual energy for capacity $P_c$: $E = 730 \sum \min(P_i, P_c)$ kWh.

Benefit–cost table

Benefit = $E \times 0.04$; Cost = $P_c \times 101.81$.

Incremental check: going from 320 to 400 kW adds 80 kW (cost $8,145) and 233,600 kWh (benefit $9,344): worth it. Going from 400 to 510 kW adds 110 kW (cost $11,199) but only 240,900 kWh (benefit $9,636): not worth it. A kW is worth installing only if it runs at least $101.81/(8760 \times 0.04) = 29\%$ of the year (about 3.5 months); the 400–510 kW band runs only 3 months.

Results

• Best installed capacity = 400 kW (maximum net benefit $43,373/yr).
• Firm energy (minimum power 65 kW available all year):

$$E_f = 65 \times 8760 = 569{,}400\ \text{kWh} = 0.569\ \text{GWh}$$

• Total energy at 400 kW = 2,102,400 kWh = 2.102 GWh.
• Secondary energy = 2,102,400 − 569,400 = 1,533,000 kWh = 1.533 GWh.
• Plant factor:

$$PF = \frac{2{,}102{,}400}{400 \times 8760} = 0.60 = 60\%$$

Answer: Installed capacity = 400 kW; firm energy = 0.569 GWh; secondary energy = 1.533 GWh; total energy = 2.102 GWh; plant factor = 60%.

The best installed capacity is the one that gives the largest net annual benefit (revenue from energy minus annualised cost of capacity). It is found by trying each power level of the power duration data as the plant capacity.

Annual cost of capacity

No O&M cost is given, so only capital recovery is used.

$$\begin{aligned} CRF &= \frac{0.12(1.12)^{50}}{(1.12)^{50}-1} = 0.12042 \\ C_a &= 750 \times 0.12042 = 90.31\ \text{\$/kW/yr} \end{aligned}$$

Energy price = 36 $/MWh = 0.036 $/kWh. Each month = 730 h.

Power duration (descending order)

700, 590, 535, 300, 255, 180, 150, 105, 80, 70, 60, 50 kW.

For capacity $P_c$: $E = 730\sum \min(P_i,P_c)$, Benefit $= 0.036E$, Cost $= 90.31P_c$.

Incremental check

• 255 → 300 kW: +45 kW costs $4,064; extra energy 131,400 kWh (4 months) earns $4,730. Worth adding.
• 300 → 535 kW: +235 kW costs $21,223; extra energy 514,650 kWh (3 months) earns $18,527. Not worth adding.

Break-even running time = $90.31/(8760\times0.036) = 28.6\%$ of the year (3.4 months). The band above 300 kW runs only 3 months, so it does not pay.

Results at 300 kW

• Firm energy $= 50 \times 8760 = 438{,}000$ kWh
• Total energy = 1,569,500 kWh; secondary energy = 1,131,500 kWh
• Plant factor $= 1{,}569{,}500/(300\times8760) = 59.7\%$

Answer: Best installed capacity = 300 kW (net benefit ≈ $29,408 per year).

Primary (firm) energy is the energy that can be produced all year from the minimum (dependable) flow. Secondary energy is the extra energy produced when flow is above the minimum, up to the plant capacity.

Step 1: Flow duration (Weibull plotting position $P = \frac{m}{N+1}\times100$, N = 12)

Step 2: Flow at 45% time (linear interpolation)

$$Q_{45} = 42 - \frac{45-38.46}{46.15-38.46}(42-39) = 39.45\ \text{m}^3/\text{s}$$

Step 3: Plant capacity

$P = \gamma Q H \eta = 9.81 \times Q \times 170 \times 0.85 = 1417.55\,Q$ kW

$$P_{inst} = 1417.55 \times 39.45 = 55{,}922\ \text{kW} \approx 55.9\ \text{MW}$$

Step 4: Primary energy (minimum flow 18 m³/s, May)

$$\begin{aligned} P_f &= 1417.55 \times 18 = 25{,}516\ \text{kW} \\ E_p &= 25{,}516 \times 8760 = 223.52\ \text{GWh} \end{aligned}$$

Step 5: Total energy (flow used = smaller of Q and 39.45; 730 h per month)

$\sum Q_{used} = 402.25$ m³/s-months.

$$\begin{aligned} E_{total} &= 1417.55 \times 402.25 \times 730 = 416.25\ \text{GWh} \\ E_s &= 416.25 - 223.52 = 192.73\ \text{GWh} \end{aligned}$$

Answer: Plant capacity ≈ 55.9 MW; primary energy ≈ 223.5 GWh/yr; secondary energy ≈ 192.7 GWh/yr (total ≈ 416.3 GWh/yr).

(If $P = m/N$ is used for the duration curve, $Q_{45} \approx 40.8$ m³/s and the secondary energy rises slightly; the method is the same.)

Primary energy is generated by the flow available 100% of the time (minimum flow). Secondary energy is the additional energy from flows above the minimum, limited by the plant (design) discharge of 55 m³/s.

Power per unit discharge

$$P = 9.81\,Q\,H\,\eta = 9.81 \times Q \times 102 \times 0.76 = 760.47\,Q\ \text{kW}$$

Installed capacity $= 760.47 \times 55 = 41{,}826$ kW ≈ 41.8 MW.

Primary energy

Minimum flow = 18 m³/s (May).

$$\begin{aligned} P_f &= 760.47 \times 18 = 13{,}688\ \text{kW} \\ E_p &= 13{,}688 \times 8760 = 119.91 \times 10^6\ \text{kWh} = 119.91\ \text{GWh} \end{aligned}$$

Monthly energy (flow used ≤ 55 m³/s, 730 h per month)

$$\begin{aligned} E_{total} &= 760.47 \times 467 \times 730 = 259.25\ \text{GWh} \\ E_s &= 259.25 - 119.91 = 139.34\ \text{GWh} \end{aligned}$$

Answer: Primary energy ≈ 119.9 GWh/yr; secondary energy ≈ 139.3 GWh/yr; total ≈ 259.3 GWh/yr.

Firm energy comes from the minimum flow, which is available throughout the year. Secondary energy is the extra energy from flows above the minimum, up to the design discharge (900 m³/s).

Power per unit discharge

$$P = 9.81 \times Q \times 100 \times 0.90 = 882.9\,Q\ \text{kW}$$

Installed capacity $= 882.9 \times 900 = 794{,}610$ kW ≈ 794.6 MW.

Firm energy

Minimum flow = 90 m³/s (March).

$$\begin{aligned} P_f &= 882.9 \times 90 = 79{,}461\ \text{kW} \\ E_f &= 79{,}461 \times 8760 = 696.08\ \text{GWh} \end{aligned}$$

Total energy (flow used = smaller of Q and 900; 730 h per month)

$$\begin{aligned} E_{total} &= 882.9 \times 6460 \times 730 = 4163.58\ \text{GWh} \\ E_s &= 4163.58 - 696.08 = 3467.50\ \text{GWh} \end{aligned}$$

Answer: Annual firm energy ≈ 696.1 GWh; annual secondary energy ≈ 3467.5 GWh (total ≈ 4163.6 GWh).

Firm energy is the energy from the minimum flow (available all year). Secondary energy is the extra energy from flows above the minimum, limited by the design discharge of 900 m³/s.

Power per unit discharge

$$P = 9.81 \times Q \times 100 \times 0.90 = 882.9\,Q\ \text{kW}$$

Installed capacity $= 882.9 \times 900 = 794.61$ MW.

Firm energy

Minimum flow = 100 m³/s (January and November).

$$\begin{aligned} P_f &= 882.9 \times 100 = 88{,}290\ \text{kW} \\ E_f &= 88{,}290 \times 8760 = 773.42\ \text{GWh} \end{aligned}$$

Total energy (730 h per month)

$$\begin{aligned} E_{total} &= 882.9 \times 5880 \times 730 = 3789.76\ \text{GWh} \\ E_s &= 3789.76 - 773.42 = 3016.34\ \text{GWh} \end{aligned}$$

Answer: Annual firm energy ≈ 773.4 GWh; annual secondary energy ≈ 3016.3 GWh (total ≈ 3789.8 GWh).

Primary energy is produced by the minimum flow that is available all year; secondary energy is produced by the flow above this minimum. No design discharge is given, so the plant is assumed to use all the river flow (secondary energy is then the maximum possible). Each month is taken as 730 h ($2.628 \times 10^6$ s).

Power per unit discharge

$$P = 9.81 \times Q \times 200 \times 0.80 = 1569.6\,Q\ \text{kW}$$

Primary energy

Minimum flow = 100 m³/s (January).

$$\begin{aligned} P_f &= 1569.6 \times 100 = 156{,}960\ \text{kW} = 156.96\ \text{MW} \\ E_p &= 156{,}960 \times 8760 = 1374.97\ \text{GWh} \end{aligned}$$

Total and secondary energy

$$E_s = 12{,}592.43 - 1374.97 = 11{,}217.46\ \text{GWh}$$

Pondage for firm capacity increased by 150%

"Increased by 150%" is read as new firm capacity = $1 + 1.5 = 2.5$ times the original.

$$\begin{aligned} P_{f,new} &= 2.5 \times 156.96 = 392.4\ \text{MW} \\ Q_{f,new} &= 2.5 \times 100 = 250\ \text{m}^3/\text{s} \end{aligned}$$

Months with flow below 250 m³/s must be supplied from storage:

$$\begin{aligned} V &= 390 \times 2.628\times10^6 \\ &= 1.025 \times 10^9\ \text{m}^3 \end{aligned}$$

This storage is easily refilled in the monsoon months, when flows are far above 250 m³/s.

Answer: Primary energy ≈ 1375.0 GWh/yr; secondary energy ≈ 11,217.5 GWh/yr; storage (pondage) required ≈ 1.025 × 10⁹ m³ (1025 million m³).

(If "increased by 150%" is taken as 1.5 times, i.e. firm flow 150 m³/s, the deficits are 50 + 30 + 10 = 90 m³/s-month and the storage is $90 \times 2.628\times10^6 = 2.37\times10^8$ m³.)

Firm energy is produced by the minimum flow (available 100% of the time); secondary energy is the extra energy from flow above the minimum.

Assumptions

The head, efficiency and plant capacity are not given, so assume:

• Net head $H = 100$ m, overall efficiency $\eta = 0.85$
• The plant can use all the river flow (no spilling)
• Each month = 730 h

Note: the June flow (100 m³/s) is taken as printed, so June is the minimum-flow month.

Power per unit discharge

$$P = 9.81 \times Q \times 100 \times 0.85 = 833.85\,Q\ \text{kW}$$

Firm energy

Minimum flow = 100 m³/s.

$$\begin{aligned} P_f &= 833.85 \times 100 = 83{,}385\ \text{kW} \\ E_f &= 83{,}385 \times 8760 = 730.45\ \text{GWh} \end{aligned}$$

Total energy

$$\begin{aligned} E_{total} &= 833.85 \times 8475 \times 730 = 5158.82\ \text{GWh} \\ E_s &= 5158.82 - 730.45 = 4428.37\ \text{GWh} \end{aligned}$$

In general form: $E_f = 9.81\,\eta H \times 100 \times 8760$ and $E_s = 9.81\,\eta H \times (8475 - 12\times100) \times 730$ kWh, so the answer scales directly with the assumed $\eta H$.

Answer (H = 100 m, η = 85%): firm energy ≈ 730.5 GWh/yr; secondary energy ≈ 4428.4 GWh/yr.

Primary energy comes from the minimum usable flow; secondary energy from the usable flow above it, up to the design discharge. An environmental flow must first be subtracted from the river flow.

Environmental flow

Minimum river flow = 100 m³/s (January).

$$Q_{env} = 0.10 \times 100 = 10\ \text{m}^3/\text{s}\ \text{(released every month)}$$

Usable flow = Q − 10, limited to the design discharge 950 m³/s.

Power per unit discharge

$$P = 9.81 \times Q \times 150 \times 0.85 = 1250.78\,Q\ \text{kW}$$

Installed capacity $= 1250.78 \times 950 = 1{,}188{,}236$ kW ≈ 1188.2 MW.

Primary energy

Minimum usable flow = 100 − 10 = 90 m³/s.

$$\begin{aligned} P_f &= 1250.78 \times 90 = 112{,}570\ \text{kW} \\ E_p &= 112{,}570 \times 8760 = 986.11\ \text{GWh} \end{aligned}$$

Total energy (730 h per month)

$$\begin{aligned} E_{total} &= 1250.78 \times 6780 \times 730 = 6190.59\ \text{GWh} \\ E_s &= 6190.59 - 986.11 = 5204.47\ \text{GWh} \end{aligned}$$

Answer: Primary energy ≈ 986.1 GWh/yr; secondary energy ≈ 5204.5 GWh/yr (installed capacity ≈ 1188 MW).

Primary energy is the energy available throughout the year from the minimum usable flow; secondary energy is the extra energy from usable flows above that minimum. No design discharge is given, so the plant is assumed to use all the usable flow. Each month is taken as 730 h (Nepali months vary from 29 to 32 days, but the yearly total is the same).

Environmental flow

Minimum flow of the series = 400 m³/s (month 10, Magh).

$$Q_{env} = 0.10 \times 400 = 40\ \text{m}^3/\text{s}$$

Usable flow = Q − 40.

Power per unit discharge

$$P = 9.81 \times Q \times 155 \times 0.80 = 1216.44\,Q\ \text{kW}$$

Primary energy

Minimum usable flow = 400 − 40 = 360 m³/s.

$$\begin{aligned} P_f &= 1216.44 \times 360 = 437{,}918\ \text{kW} \approx 437.9\ \text{MW} \\ E_p &= 437{,}918 \times 8760 = 3836.17\ \text{GWh} \end{aligned}$$

Total energy

$$\begin{aligned} E_{total} &= 1216.44 \times 13{,}180 \times 730 = 11{,}703.86\ \text{GWh} \\ E_s &= 11{,}703.86 - 3836.17 = 7867.69\ \text{GWh} \end{aligned}$$

Answer: Primary energy ≈ 3836.2 GWh/yr; secondary energy ≈ 7867.7 GWh/yr (total ≈ 11,703.9 GWh/yr).

A power duration curve is the flow duration curve with the flow axis converted to power, $P = 9.81\,Q H \eta$. Area under it gives energy.

Power for each flow

$$P = 9.81 \times Q \times 100 \times 0.89 = 873.09\,Q\ \text{kW}$$

P (MW)
 786 |*
     | \
 524 |   *
 437 |     *
 349 |========*======  <- capacity 331.8 MW
 314 |        |  *  *         (45% time)
 262 |        |        *  *
 175 |  secondary          *
 122 |        |               *
  87 |-primary (firm)---------------*
     +---------------------------------
     0   20   45   60   80   93  100  % time

Plant capacity (flow at 45% time)

$$\begin{aligned} Q_{45} &= 400 - \frac{45-40}{50-40}(400-360) = 380\ \text{m}^3/\text{s} \\ P_{inst} &= 873.09 \times 380 = 331{,}774\ \text{kW} \approx 331.8\ \text{MW} \end{aligned}$$

Primary energy

Flow available 100% of the time = 100 m³/s.

$$\begin{aligned} P_f &= 873.09 \times 100 = 87{,}309\ \text{kW} \\ E_p &= 87{,}309 \times 8760 = 764.83\ \text{GWh} \end{aligned}$$

Total energy (area under capped curve, trapezoidal rule)

Flow used is capped at 380 m³/s from 0% to 45%.

Mean flow used $= 32{,}300/100 = 323$ m³/s.

$$\begin{aligned} E_{total} &= 873.09 \times 323 \times 8760 = 2470.39\ \text{GWh} \\ E_s &= 2470.39 - 764.83 = 1705.56\ \text{GWh} \end{aligned}$$

Plant factor $= 323/380 = 85\%$.

Answer: Installed capacity ≈ 331.8 MW; primary energy ≈ 764.8 GWh/yr; secondary energy ≈ 1705.6 GWh/yr.

Firm and secondary power

• Firm (primary) power is the power that is available continuously (100% of the time), even in the driest period. It is fixed by the minimum flow.
• Secondary (surplus) power is the power above firm power that is available only part of the time, when flow is higher than the minimum. It is sold at a lower rate or used for non-essential loads.

Data

Minimum load = 12,000 kW; maximum load = 35,000 kW; two units of 22,000 kW each. The load is assumed to vary linearly (uniformly) between minimum and maximum, so the average load is the mean of the two.

$$P_{avg} = \frac{12{,}000 + 35{,}000}{2} = 23{,}500\ \text{kW}$$

(i) Total installed capacity

$$P_{inst} = 2 \times 22{,}000 = 44{,}000\ \text{kW}$$

(ii) Plant factor (capacity factor)

$$PF = \frac{\text{average load}}{\text{installed capacity}} = \frac{23{,}500}{44{,}000} = 0.534 = 53.4\%$$

(iii) Maximum demand

Maximum demand = peak load = 35,000 kW.

(iv) Load factor

$$LF = \frac{\text{average load}}{\text{maximum demand}} = \frac{23{,}500}{35{,}000} = 0.671 = 67.1\%$$

(v) Utilization factor

$$UF = \frac{\text{maximum demand}}{\text{installed capacity}} = \frac{35{,}000}{44{,}000} = 0.795 = 79.5\%$$

Check: $PF = LF \times UF = 0.671 \times 0.795 = 0.534$.

Answer: 44,000 kW; 53.4%; 35,000 kW; 67.1%; 79.5%.

Flow duration curve (FDC)

A flow duration curve is a plot of discharge (on the y-axis) against the percentage of time that discharge is equalled or exceeded (on the x-axis). It is a cumulative frequency curve of the river flow.

How it is drawn

1. Collect a long record of daily or monthly flows.
2. Arrange the flows in descending order and give each a rank $m$ (1 for the largest).
3. Compute the exceedance probability $P = \frac{m}{N+1}\times100\%$ (N = number of values).
4. Plot Q against P.

Power duration curve (PDC)

A power duration curve is obtained by converting each flow of the FDC to power:

$$P = 9.81\,Q\,H\,\eta\ \text{kW}$$

It shows the percentage of time a given power is equalled or exceeded. Where the head is nearly constant, the PDC has the same shape as the FDC with a different scale.

Q or P
  |*
  | *
  |  **
  |    ***  installed capacity
  |=======*****======   (e.g. Q40)
  |   secondary  ****
  |                 *****
  |--- firm (Q100) -------****
  +---------------------------->
  0        50             100
       % time equalled/exceeded

Significance in estimating power potential

• Firm power: the power at 100% (or 90–95% in practice) of time gives the firm (dependable) power of a run-of-river plant.
• Installed capacity selection: the design discharge is chosen at a certain exceedance, e.g. Q40 to Q65 for Nepalese run-of-river projects; the PDC shows the power at that level.
• Energy estimation: the area under the PDC (capped at the installed capacity) gives the annual energy; the area below firm power is primary energy and the rest is secondary energy.
• Plant factor: mean power from the capped curve divided by installed capacity.
• Flow character: a steep curve shows a flashy river (needs storage); a flat curve shows a well-regulated, groundwater-fed river.
• Storage and pondage: shows how much storage is needed to raise the firm flow.
• Environmental flow and spill: shows how often water is spilled and how much is left for downstream use.

Flow duration curve

A flow duration curve (FDC) is a graph of river discharge against the percentage of time that discharge is equalled or exceeded. It is drawn by arranging the recorded flows in descending order and plotting each against $P = \frac{m}{N+1}\times 100\%$. It is used to find firm flow, design discharge (e.g. Q40, Q65) and annual energy.

Parameters influencing planning of hydropower projects

1. Hydrology: the amount and variation of river flow (FDC, mean monthly flows, floods, low flows) decides firm power, installed capacity, and spillway size. Long, reliable records are needed.
2. Topography (head): power depends on $Q \times H$. Steep rivers with a short distance between intake and powerhouse give high head cheaply. Valley shape decides dam type, reservoir area and layout.
3. Geology: the foundation of the dam, tunnel route and powerhouse must be strong and stable. Faults, weak rock, landslides and seismic zones raise cost and risk (very important in the young Himalaya).
4. Sediment: Himalayan rivers carry heavy sediment, which needs desanders and causes turbine wear and reservoir filling.
5. Power market and demand: size and type of plant (base load or peaking, run-of-river or storage) must match the load curve, demand growth and possibility of export.
6. Access and transmission: distance to roads and to the grid; long transmission lines add cost and losses.
7. Environmental and social factors: land acquisition, resettlement, environmental flow, fish, effect on downstream users; an EIA/IEE is required.
8. Economic and financial factors: project cost, interest rate, energy tariff, benefit–cost ratio, IRR and payback period.
9. Legal and institutional aspects: licences, water rights, government policy and royalty.
10. Construction materials and logistics: availability of sand, aggregate and cement near the site.
11. Multipurpose use: irrigation, flood control, water supply and navigation needs can change the reservoir operation.

Pre-feasibility study is a preliminary study that compares alternative layouts of a hydropower site and selects the most promising one. Feasibility study is a detailed study of the selected scheme to prove that it is technically, economically, financially and environmentally viable and ready for financing.

In short, the pre-feasibility study answers "which option is best?", and the feasibility study answers "is this option worth building, and how?".

Hydropower planning depends on several parameters. Three of the most important are the power market, hydrology and geology.

Power market

The power market is the demand for the energy and the price it can be sold at.

• Load demand and growth: the plant size must match present and future demand; an oversized plant spills energy.
• Load pattern: base-load demand suits run-of-river plants; peak demand needs peaking run-of-river (PROR) or storage plants.
• Seasonal mismatch: in Nepal, run-of-river plants have surplus energy in the monsoon and a shortage in winter, so storage or export markets (India, Bangladesh) are important.
• Tariff and power purchase agreement (PPA): the buying rate (e.g. NEA rates, different for wet and dry season) decides revenue and financial viability.
• Grid access: distance to the grid and transmission capacity.

Hydrology

Hydrology tells how much water is available and when.

• Mean monthly flows and FDC: fix firm flow, design discharge (e.g. Q40–Q65) and annual energy.
• Flood flows: design flood (e.g. 1-in-1000 or 10,000-year) sizes the spillway, diversion works and cofferdams.
• Low flows: decide firm power and environmental flow.
• Sediment load: sizes the desander and estimates reservoir life.
• Glacial lakes (GLOF), snowmelt, climate change: need special study in Himalayan basins.
• Long, reliable records are needed; for ungauged rivers, regional methods (e.g. MIP, WECS/DHM methods) are used.

Geology

Geology decides whether the structures can be safely and economically built.

• Foundation: dam and powerhouse need strong, watertight rock with low permeability.
• Tunnelling conditions: rock class, faults, shear zones, squeezing and water ingress control tunnel cost and time.
• Slope stability: landslides at the intake, headworks and penstock route.
• Seismicity: Nepal is a high seismic zone; design must include earthquake loads.
• Construction materials: sand, aggregate, clay for dams.
• Reservoir leakage and sedimentation depend on geology of the catchment.

Weak geology is the most common reason for cost and time overruns of Nepalese hydropower projects.

Hydro-electric power is electricity produced by using the potential energy of water falling through a head to drive turbines coupled to generators ($P = 9.81QH\eta$).

Advantages

1. Renewable source: water is renewed by the hydrological cycle and is not used up.
2. No fuel cost: running cost is very low; only operation and maintenance.
3. Clean: no smoke, ash or CO₂ during operation.
4. High efficiency: overall efficiency 80–90%, much higher than thermal plants (30–40%).
5. Quick start and load following: units can start and take load in a few minutes, so they are ideal for peak load and frequency control.
6. Long life: 50–100 years for civil works.
7. Reliable and simple: fewer moving parts, low breakdown rate.
8. Multipurpose benefits: storage projects also provide irrigation, flood control, water supply, fisheries and tourism.
9. No import dependence: for Nepal it uses local resources and saves foreign currency; surplus can be exported.

Limitations

1. High initial cost: dams, tunnels and access roads are costly.
2. Long construction period: usually 4–10 years.
3. Dependent on rainfall/flow: output falls in the dry season or drought years.
4. Site specific: plants are in remote hills, far from load centres, needing long transmission lines.
5. Environmental and social effects: submergence of land, displacement of people, effect on fish and the downstream river.
6. Sedimentation: reservoirs fill up and turbines are eroded, especially in Himalayan rivers.
7. Geological and seismic risk: landslides, weak rock and earthquakes can damage works.
8. Risk of dam failure and GLOF: large consequences downstream.

Major sources of electric power generation

• Conventional: hydropower, thermal (coal, oil, diesel), gas turbine and combined cycle, nuclear.
• Non-conventional (renewable): solar (photovoltaic and solar thermal), wind, biomass and biogas, geothermal, tidal and wave energy, fuel cells.

Hydropower compared with thermal power

Merits of hydropower: no fuel cost, clean, high efficiency, long life, quick start, multipurpose use.

Demerits of hydropower: high capital cost and long construction time, output varies with season, remote sites need long transmission lines, and dams cause submergence and resettlement.