BE Civil Engineering (IOE, TU) Hydropower Engineering (IOE, CE 755) Question Paper 2080 Nepal

(a) Flow-duration curve and 8-month dependable discharge

Rank the 12 monthly discharges in descending order and assign exceedance percentage $P = \dfrac{m}{N+1}\times100$ where $N=12$:

The flow that is equalled or exceeded for at least 8 of the 12 months is the 8th-ranked value, i.e. the discharge available 8 months/year:

$$Q_{8} = 20\ \text{m}^3/\text{s} \quad (\approx 61.5\% \text{ exceedance} \approx 8/12 = 67\%)$$

(b) Firm power

$$P = \rho\, g\, Q\, H\, \eta$$ $$P = 1000 \times 9.81 \times 20 \times 120 \times 0.82$$ $$P = 9.81\times20 = 196.2;\ \times120 = 23{,}544;\ \times0.82 = 19{,}306.08\ \text{kW}$$ $$\boxed{P_{firm} \approx 19.31\ \text{MW}}$$

(c) Average annual energy (turbine capped at $Q_d = 45$ m³/s)

Usable monthly discharge $Q_u = \min(Q_{month}, 45)$:

Sum of usable discharges $= 18+14+12+11+20+45+45+45+45+40+28+22 = 345\ \text{m}^3/\text{s}$.

Mean usable discharge $\bar Q_u = 345/12 = 28.75\ \text{m}^3/\text{s}$.

Average power:

$$\bar P = \rho g \bar Q_u H \eta = 1000\times9.81\times28.75\times120\times0.82$$ $$= 9.81\times28.75 = 282.0375;\ \times120 = 33{,}844.5;\ \times0.82 = 27{,}752.49\ \text{kW} \approx 27.75\ \text{MW}$$

Annual energy (8760 h/yr):

$$E = \bar P \times 8760 = 27{,}752.49\ \text{kW} \times 8760\ \text{h} = 243{,}11\times10^{3}\ \text{kWh}$$ $$E = 2.4311\times10^{8}\ \text{kWh} = 243.1\ \text{GWh}$$ $$\boxed{E \approx 243\ \text{GWh/year}}$$

(d) Nepal's rivers are snow- and monsoon-fed: roughly 80 % of annual flow occurs in the wet months (Jun–Sep). In the dry season flows drop to a small fraction of the peak (here 11 vs 95 m³/s, a ~9:1 ratio). A run-of-river plant has negligible storage, so generation tracks the natural hydrograph — full output during monsoon and only firm (low) output in winter. This is why RoR projects sell large secondary (wet-season) energy but provide modest firm power, and why peaking-reserve / storage projects are needed for dry-season security.

(a) Fall velocity of a 0.30 mm particle

Stokes' law (laminar, valid for $Re_p < 1$):

$$w = \frac{g\,(s-1)\,d^2}{18\,\nu}$$

with $g = 9.81$, $s = 2.65$, $d = 0.30\times10^{-3}\ \text{m}$, $\nu = 1.0\times10^{-6}$:

$$w = \frac{9.81\,(1.65)\,(0.30\times10^{-3})^2}{18\times1.0\times10^{-6}} = \frac{9.81\times1.65\times9.0\times10^{-8}}{1.8\times10^{-5}}$$ $$= \frac{1.4568\times10^{-6}}{1.8\times10^{-5}} = 0.0809\ \text{m/s}$$

Check particle Reynolds number:

$$Re_p = \frac{w\,d}{\nu} = \frac{0.0809\times0.30\times10^{-3}}{1.0\times10^{-6}} = 24.3$$

Since $Re_p = 24.3 \gg 1$, Stokes' law is not valid (we are in the transition range). Use a transition-range estimate. A commonly adopted design value for a 0.30 mm sand grain (the empirical settling chart / Sudry / Scobey curves) is:

$$\boxed{w \approx 0.035\ \text{m/s}}$$

(Stokes badly over-predicts because form drag is neglected. Equivalently the transition drag relation $C_D \approx 18.5\,Re_p^{-0.6}$ gives $w$ on the order of 0.03–0.04 m/s for this grain size.)

(b) Basin dimensions (ideal settling)

Water must travel down depth $D$ in the same time it travels along length $L$:

$$\frac{D}{w} = \frac{L}{V} \;\Rightarrow\; L = \frac{V\,D}{w}$$ $$L = \frac{0.25\times3.5}{0.035} = \frac{0.875}{0.035} = 25.0\ \text{m}$$

Required flow cross-section (continuity $Q = V\,A = V\,B\,D$):

$$B = \frac{Q}{V\,D} = \frac{12}{0.25\times3.5} = \frac{12}{0.875} = 13.7\ \text{m}$$ $$\boxed{L \approx 25\ \text{m}, \quad B \approx 13.7\ \text{m}, \quad D = 3.5\ \text{m}}$$

(c) Turbulence correction

Real flow is turbulent, so an effective (reduced) fall velocity $w' = w - \Delta w$ is used. A simple Camp/Vetter correction multiplies length by a factor $\lambda$ (typically 1.2–1.5). Taking $\lambda = 1.4$:

$$L_{design} = 1.4 \times 25.0 = 35\ \text{m}$$

Thus provide $L \approx 35\ \text{m}$, $B \approx 13.7\ \text{m}$ (often split into 2–3 parallel chambers ≈ 5 m wide each for hydraulic stability and flushing). The basin should include a transition zone, a sloped flushing channel, and a periodic/continuous flushing arrangement, because Himalayan rivers carry very high sediment loads, especially abrasive quartz, that would otherwise erode the turbine runners.

(a) Maximum upsurge (frictionless)

Tunnel velocity at design flow:

$$V_0 = \frac{Q}{A_t} = \frac{18}{9.0} = 2.0\ \text{m/s}$$

For a simple surge tank with no friction, energy conservation between the kinetic energy of the tunnel water column and the potential energy of the raised water in the tank gives the maximum surge amplitude:

$$Z_{max} = V_0\sqrt{\frac{L\,A_t}{g\,A_s}}$$ $$\frac{L\,A_t}{g\,A_s} = \frac{800\times9.0}{9.81\times40} = \frac{7200}{392.4} = 18.349\ \text{s}^2$$ $$\sqrt{18.349} = 4.283\ \text{s}$$ $$Z_{max} = 2.0\times4.283 = 8.57\ \text{m}$$ $$\boxed{Z_{max} \approx 8.57\ \text{m above static level}}$$

(b) Period of mass oscillation

The undamped (frictionless) mass oscillation is simple harmonic with period:

$$T = 2\pi\sqrt{\frac{L\,A_s}{g\,A_t}}$$ $$\frac{L\,A_s}{g\,A_t} = \frac{800\times40}{9.81\times9.0} = \frac{32000}{88.29} = 362.44\ \text{s}^2$$ $$\sqrt{362.44} = 19.038\ \text{s}$$ $$T = 2\pi\times19.038 = 119.6\ \text{s}$$ $$\boxed{T \approx 120\ \text{s} \;(\approx 2.0\ \text{min})}$$

(c) Function and choice

A surge tank is a free water surface installed between the headrace tunnel and the penstock. Its functions are: (i) to protect the long pressure tunnel from water-hammer by reflecting the pressure waves (it converts a sudden flow change into a slow mass oscillation), (ii) to provide a ready supply / storage of water to the turbine during load demand increases, and (iii) to improve speed regulation.

A restricted-orifice (throttled) surge tank is preferred where it is desirable to introduce additional damping and to develop a retarding head quickly: the orifice creates a head loss that damps oscillations faster and reduces the tank size and surge amplitude, useful for plants with frequent load changes or where a tall simple tank would be uneconomic.

(a) Penstock diameter

From continuity $Q = V\,A = V\,\dfrac{\pi}{4}D^2$, with target $V \approx 4.0$ m/s:

$$D = \sqrt{\frac{4Q}{\pi V}} = \sqrt{\frac{4\times18}{\pi\times4.0}} = \sqrt{\frac{72}{12.566}} = \sqrt{5.730} = 2.394\ \text{m}$$

Adopt a standard diameter $\boxed{D = 2.4\ \text{m}}$.

(b) Actual velocity and Joukowsky pressure rise

Area $A = \dfrac{\pi}{4}(2.4)^2 = 0.7854\times5.76 = 4.5239\ \text{m}^2$.

$$V = \frac{Q}{A} = \frac{18}{4.5239} = 3.979\ \text{m/s}$$

Joukowsky (instantaneous closure) pressure rise:

$$\Delta p = \rho\,a\,V \quad\Rightarrow\quad \Delta H = \frac{a\,V}{g}$$ $$\Delta H = \frac{1000\times3.979}{9.81} = \frac{3979}{9.81} = 405.6\ \text{m}$$

As a pressure: $\Delta p = \rho a V = 1000\times1000\times3.979 = 3.979\times10^{6}\ \text{Pa} = 3.98\ \text{MPa} \approx 39.8\ \text{bar}$.

As a percentage of net head:

$$\frac{\Delta H}{H} = \frac{405.6}{250} = 1.62 = 162\%$$ $$\boxed{\Delta H \approx 406\ \text{m} \;(\approx 162\% \text{ of net head})}$$

This very large surge (the total transient head reaches $250 + 406 = 656$ m) shows why instantaneous closure must never be allowed; gate closure is slowed and/or a surge tank/relief valve provided.

(c) Critical closure time and criterion

$$T_c = \frac{2L}{a} = \frac{2\times450}{1000} = 0.90\ \text{s}$$

Criterion:

• If the gate closing time $T_{close} \le T_c = 2L/a$, the closure is sudden (rapid); the full Joukowsky head $\dfrac{aV}{g}$ develops and the pipe must be designed for it.
• If $T_{close} > 2L/a$, the closure is gradual (slow); the pressure rise is smaller and may be estimated by formulae such as Allievi's, since the returning negative wave partially relieves the pressure before closure completes.

(a) Power

Total hydraulic-to-shaft power available:

$$P_{total} = \rho g Q H \eta = 1000\times9.81\times18\times250\times0.90$$ $$= 9.81\times18 = 176.58;\ \times250 = 44{,}145;\ \times0.90 = 39{,}730.5\ \text{kW} \approx 39.73\ \text{MW}$$

With two identical units, power per unit:

$$P_{unit} = \frac{39{,}730.5}{2} = 19{,}865\ \text{kW} \approx 19.87\ \text{MW}$$ $$\boxed{P_{unit}\approx19.87\ \text{MW},\quad P_{total}\approx39.73\ \text{MW}}$$

(b) Turbine selection and synchronous speed

A net head of 250 m is a high head. Reaction turbines (Francis) are typically used up to ~300–350 m, and Pelton impulse turbines for very high heads (>250–300 m). At 250 m both are candidates; for a high-head, moderate-discharge site with abrasive Himalayan sediment, a Pelton turbine (or a high-head Francis) is appropriate. We select a Pelton turbine because sediment-laden water at high head favours the more easily maintained/replaceable impulse runner and buckets.

Synchronous speed: $N = \dfrac{120 f}{p}$ with $f = 50$ Hz. Choosing $p = 12$ poles:

$$N = \frac{120\times50}{12} = 500\ \text{rpm}$$

(c) Specific speed (power-based, metric)

$$N_s = \frac{N\sqrt{P}}{H^{5/4}}\qquad (P\text{ in kW},\ H\text{ in m},\ N\text{ in rpm})$$

For one unit, $P = 19{,}865$ kW:

$$\sqrt{P} = \sqrt{19865} = 140.9$$ $$H^{5/4} = 250^{1.25}: \ln 250 = 5.5215,\ \times1.25 = 6.9019,\ e^{6.9019} = 994.0$$ $$N_s = \frac{500\times140.9}{994.0} = \frac{70{,}450}{994.0} = 70.9$$ $$\boxed{N_s \approx 71\ \text{(metric, single-jet basis)}}$$

For a single-jet Pelton the recommended $N_s$ range is roughly 10–35; $N_s \approx 71$ is too high for one jet, which means a multi-jet Pelton is implied. Using a Pelton with $n$ jets, the per-jet specific speed is $N_{s,jet}=N_s/\sqrt{n}$. With $n = 4$ jets:

$$N_{s,jet} = \frac{71}{\sqrt{4}} = 35.5$$

which falls at the upper edge of the Pelton range, confirming a 4-jet vertical Pelton unit at 500 rpm is feasible. (Alternatively, $N_s\approx71$ lies squarely in the medium-head Francis range of about 60–150, so a Francis turbine at 250 m is equally defensible — either selection is accepted if justified.)

(i) Classification by head

• Low head: $H < 30\ \text{m}$ — usually large discharge, Kaplan/propeller turbines.
• Medium head: $30\ \text{m} \le H \le 250\ \text{m}$ — Francis turbines typical.
• High head: $H > 250\ \text{m}$ — Pelton (impulse) turbines, small discharge. (Different texts use slightly different cut-offs, e.g. low <15 m, high >70/300 m.)

(ii) Classification by way of operation / water use

• Run-of-river (RoR): no/negligible storage; generation follows river flow (e.g. Marsyangdi, Kaligandaki-A in Nepal are large RoR/peaking-RoR plants).
• Storage (reservoir) plants: a dam stores wet-season flow for dry-season/peak use (e.g. Kulekhani — Nepal's only major reservoir plant).
• Pumped-storage plants: pump water to an upper reservoir during off-peak, generate during peak.
• Peaking RoR (PRoR): small pondage to meet daily peak.

(iii) Classification by capacity (Nepal convention)

• Micro: up to 100 kW.
• Mini: 100 kW – 1 MW.
• Small: 1 – 25 MW (sometimes up to 10 MW in older definitions).
• Medium: 25 – 300 MW.
• Large: > 300 MW (e.g. Upper Tamakoshi, 456 MW).

Nepali examples: Kulekhani I (storage, high head ~550 m, 60 MW); Upper Tamakoshi (RoR/peaking, 456 MW); micro-hydro schemes (<100 kW) widely used for rural electrification in hill districts.

Forebay vs surge tank

When provided:

• A forebay is provided when the headrace is an open canal (low-pressure, run-of-river hill schemes).
• A surge tank is provided when the headrace is a long pressure tunnel/pipe so that water-hammer must be controlled and a free surface created close to the turbine.

Forebay spillway and freeboard:

• The spillway safely discharges the excess water that arrives when the turbine load is suddenly reduced/rejected (the canal keeps bringing flow but the penstock takes less), preventing overtopping and protecting the structure.
• The freeboard is the vertical margin between the normal/maximum water level and the top of the forebay walls; it accommodates surges, wave action and the temporary rise on load rejection so the forebay does not overflow uncontrollably.

Maximum suction height

Thoma's cavitation coefficient:

$$\sigma = \frac{H_a - H_v - H_s}{H}$$

For cavitation-free operation $\sigma \ge \sigma_c$, so the limiting (maximum) suction head is when $\sigma = \sigma_c$:

$$H_s = (H_a - H_v) - \sigma_c\,H$$ $$H_s = (10.3 - 0.24) - 0.12\times80$$ $$H_s = 10.06 - 9.60 = 0.46\ \text{m}$$ $$\boxed{H_{s,max} \approx 0.46\ \text{m above tailwater}}$$

The value is small and positive, so the runner may be set only about 0.46 m above the tailwater. In practice, to provide a margin against cavitation it is safer to set the runner at or slightly below tailwater level (negative $H_s$), which increases $\sigma$ and the safety margin.

Role of the draft tube

The draft tube is a gradually diverging conduit connecting the runner exit to the tailrace. Its functions are:

1. Recovers kinetic energy of the water leaving the runner by converting velocity head into pressure head (its diverging shape decelerates the flow), thereby increasing the effective head/efficiency.
2. Recovers the suction (draft) head $H_s$, allowing the turbine to be installed above tailwater level while still using the full head down to tailrace.
3. By creating sub-atmospheric pressure at the runner outlet it must be designed so that the pressure does not fall to the vapour pressure — hence the cavitation (Thoma) limit above governs the permissible setting.

(a) Plant capacity factor

Maximum possible annual energy at full capacity:

$$E_{max} = 20{,}000\ \text{kW}\times8760\ \text{h} = 1.752\times10^{8}\ \text{kWh} = 175.2\ \text{GWh}$$

Capacity factor:

$$CF = \frac{110}{175.2} = 0.628$$ $$\boxed{CF \approx 62.8\%}$$

(b) Annual cost and unit cost

Annualised capital cost $= \text{CRF}\times\text{capital} = 0.10\times4.0\times10^{9} = 0.40\times10^{9} = \text{NRs. }400\ \text{million}$.

Annual O&M $= 2.5\%\times4.0\times10^{9} = 0.025\times4.0\times10^{9} = 0.10\times10^{9} = \text{NRs. }100\ \text{million}$.

Total annual cost:

$$C_{annual} = 400 + 100 = \text{NRs. }500\ \text{million} = 5.0\times10^{8}\ \text{NRs.}$$

Unit generation cost:

$$c = \frac{C_{annual}}{E} = \frac{5.0\times10^{8}}{110\times10^{6}\ \text{kWh}} = 4.545\ \text{NRs./kWh}$$ $$\boxed{c \approx \text{NRs. }4.55\ \text{per kWh}}$$

(c) Benefit–cost ratio

Annual revenue (benefit):

$$B = 8.0\ \text{NRs./kWh}\times110\times10^{6}\ \text{kWh} = 8.8\times10^{8} = \text{NRs. }880\ \text{million}$$

Benefit–cost ratio:

$$B/C = \frac{880}{500} = 1.76$$ $$\boxed{B/C \approx 1.76}$$

Comment: Since $B/C = 1.76 > 1$ (equivalently the selling price NRs. 8.0/kWh comfortably exceeds the generation cost NRs. 4.55/kWh), the project is economically viable. The healthy margin gives room for financing risk, hydrological uncertainty and tariff fluctuation.

Types of intakes

• Run-of-river / side (lateral) intake: water drawn from the side of a river or weir pool; common for RoR hill schemes in Nepal.
• Frontal / canal intake: flow taken frontally into a canal.
• Reservoir (dam) intake: intake tower or sluice located in the dam/reservoir, often with multiple levels for varying drawdown.
• Tyrolean / bottom (drop / trench) intake: a bottom rack across the stream bed, suited to steep, sediment- and boulder-laden mountain streams (widely used in Nepal).
• Shaft / tower intake: vertical shaft intake for deep reservoirs.

Essential requirements of a good intake

1. Admit the required design discharge with minimum head loss.
2. Exclude floating debris (trash rack) and bed load / coarse sediment.
3. Provide control (gates/stoplogs) to regulate or stop flow for maintenance.
4. Be located to avoid sediment deposition and air entrainment (adequate submergence to prevent vortices).
5. Be structurally stable and accessible for cleaning/flushing.

Trash-rack area blocked by bars

Consider one bar pitch (one bar + one clear opening) on the rack plane:

$$\text{pitch} = \text{bar thickness} + \text{clear spacing} = 12 + 60 = 72\ \text{mm}$$

Fraction of plane occupied by bar metal:

$$\frac{12}{72} = 0.1667 = 16.7\%$$ $$\boxed{\text{≈ }16.7\%\text{ of the gross rack area is blocked by bars}}$$

so the net (effective) open area is about 83.3 % of the gross rack area (before any debris clogging). The $70°$ inclination means the rack's gross area normal-to-flow is larger than the flow cross-section by a factor $1/\sin70° \approx 1.064$, which is why inclined racks are used — they ease raking and reduce approach-velocity head loss.

(a) Hydropower potential of Nepal

• Theoretical potential: about 83,000 MW (≈ 83 GW), based on the classic study of the runoff and head of Nepal's ~6,000 rivers and rivulets.
• Technically feasible potential: about 42,000 MW (≈ 42 GW).
• Economically feasible potential: commonly quoted as about 42,000 MW as well in older studies, with conservative figures around ~43,000 MW technically and a large fraction economically attractive; the installed capacity today is only a small fraction of this (a few thousand MW), so Nepal has developed only a small percentage of its feasible potential.

(b) Why most plants are run-of-river (RoR)

• Lower capital cost and shorter construction time than storage dams.
• Less land submergence, fewer resettlement and environmental issues.
• Suitable steep topography gives high head with small structures.
• Storage (reservoir) projects need large dams, big investment and complex environmental/social clearances, so very few exist (Kulekhani being the notable exception). The downside is poor dry-season firm power and large seasonal energy variation.

(c) Major challenges

1. Seasonal flow / dry-season deficit: RoR-dominated system produces surplus in monsoon and shortfall in winter, forcing imports; need for storage and reservoir projects.
2. High sediment load and geological/seismic risk: abrasive Himalayan sediment erodes turbines; landslides, GLOFs and earthquakes threaten structures and access roads.
3. Financing, transmission and market constraints: large capital needs, limited domestic capital market, evacuation (transmission line) bottlenecks, and need for cross-border power trade (e.g. export to India/Bangladesh) and reliable PPAs. (Other valid points: difficult terrain/access, governance and licensing delays, environmental flow requirements.)

(d) Recent landmark projects

• Upper Tamakoshi (456 MW): the country's largest plant (RoR/peaking), a flagship domestically funded project.
• Kulekhani (I & II, ~92 MW total): Nepal's only major reservoir/storage scheme, providing valuable peaking and dry-season energy. (Also acceptable: Kaligandaki-A 144 MW, Middle Marsyangdi 70 MW, Tanahu/Budhi Gandaki storage projects under development.)