BE Civil Engineering (IOE, TU) Hydropower Engineering (IOE, CE 755) Question Paper 2077 Nepal

(a) Definitions and significance

• Firm (primary) power is the power that can be guaranteed to be available for essentially the whole year (commonly taken at a high exceedance flow such as 90–95%). It corresponds to the dependable flow and is the marketable, reliable base load of the plant.
• Secondary (surplus) power is the additional power available only during periods of higher-than-firm flow (the wet season). It is non-guaranteed and is usually sold at a lower tariff or used for non-essential loads.
• Significance of the FDC: The flow-duration curve shows the percentage of time a given discharge is equalled or exceeded. The area under the curve represents the total volume / energy potential. Choosing the design discharge at a low exceedance (e.g. 30–50%) gives high installed capacity but lower plant factor and much spilled water; choosing a higher exceedance gives a smaller, more fully-utilised plant. It therefore lets the designer trade installed capacity against firmness of supply, spill and economics.

(b) Installed capacity and firm power

Net head: $H_n = H_g(1-0.05) = 120 \times 0.95 = 114\ \text{m}$.

Power equation: $P = \rho g Q H_n \eta_o = 9.81 \, Q H_n \eta_o\ \text{(kW, with }\rho g \approx 9.81\,\text{kN/m}^3).$

Installed capacity (design flow $Q_d = 48\ \text{m}^3/\text{s}$ at 40% exceedance):

$$P_{inst} = 9.81 \times 48 \times 114 \times 0.82 = 44\,019\ \text{kW} \approx \textbf{44.0 MW}.$$

Step check: $9.81\times48 = 470.88$; $\times114 = 53\,680.3$; $\times0.82 = 44\,017.8\ \text{kW}$.

Firm power (90% exceedance flow $Q_{90} = 16\ \text{m}^3/\text{s}$):

$$P_{firm} = 9.81 \times 16 \times 114 \times 0.82 = 14\,672\ \text{kW} \approx \textbf{14.7 MW}.$$

Step check: $9.81\times16 = 156.96$; $\times114 = 17\,893.4$; $\times0.82 = 14\,672.6\ \text{kW}$.

(c) Annual energy at 60% exceedance flow ($Q_{60}=33\ \text{m}^3/\text{s}$) and capacity factor

Representative power: $P_{60} = 9.81 \times 33 \times 114 \times 0.82 = 30\,262\ \text{kW} \approx 30.26\ \text{MW}.$ Step check: $9.81\times33 = 323.73$; $\times114 = 36\,905.2$; $\times0.82 = 30\,262.3\ \text{kW}$.

Hours per year: $8760\ \text{h}$.

$$E = 30\,262.3\ \text{kW} \times 8760\ \text{h} = 2.651\times10^8\ \text{kWh} \approx \textbf{265.1 GWh}.$$

Capacity (plant) factor:

$$CF = \frac{P_{60}}{P_{inst}} = \frac{30\,262}{44\,018} = 0.687 \approx \textbf{68.7\%}.$$

(Equivalently $E_{actual}/E_{rated} = 265.1/(44.018\times8.76) = 265.1/385.6 = 0.687$.)

The relatively high plant factor reflects designing at the 40% exceedance flow rather than the peak flow.

(a) Need and zones

Himalayan rivers carry very high suspended sediment, much of it hard quartz and feldspar. If not removed, particles $\gtrsim 0.2$–$0.3\ \text{mm}$ cause severe abrasion of penstock, valves and especially turbine runners (Pelton needles/buckets, Francis runners), reducing efficiency and life. A settling basin reduces the through-flow velocity so that target particles settle out before reaching the headrace/turbine.

Three functional zones:

1. Inlet (transition / inlet zone) – flow spreads gradually from the inlet channel to the full basin width to give uniform, low velocity.
2. Settling (working) zone – the main length where particles fall to the bed.
3. Outlet zone – clarified water is collected (often by an overflow weir) and conveyed to the headrace; sediment collects in a hopper/flushing zone at the bottom for periodic flushing.

(b) Ideal (Camp) design

In the ideal basin a particle is just removed if its settling time equals the flow-through time. The governing relation is:

$$\frac{w}{v} = \frac{D}{L} \quad\Rightarrow\quad L = \frac{vD}{w}, \qquad \text{and surface-load: } w = \frac{Q}{A_{surface}} = \frac{Q}{B L}.$$

Required surface area from overflow-rate criterion:

$$A = \frac{Q}{w} = \frac{12}{0.035} = 342.9\ \text{m}^2 \approx \textbf{343 m}^2.$$

Width from continuity (cross-section $= B D$, velocity $v$):

$$B D v = Q \Rightarrow B = \frac{Q}{D v} = \frac{12}{3.0 \times 0.20} = \frac{12}{0.60} = \textbf{20.0 m}.$$

Length (from settling criterion $L = vD/w$):

$$L = \frac{0.20 \times 3.0}{0.035} = \frac{0.60}{0.035} = 17.14\ \text{m} \approx \textbf{17.2 m}.$$

Check: $A = B L = 20.0 \times 17.14 = 342.9\ \text{m}^2$ — consistent with the overflow-rate area. ✔

(c) Turbulence / short-circuiting correction

Real basins have turbulence and short-circuiting, so the effective fall velocity is reduced. Using $w' = 0.80\,w = 0.80 \times 0.035 = 0.028\ \text{m/s}$:

$$L' = \frac{vD}{w'} = \frac{0.60}{0.028} = 21.4\ \text{m}.$$

Alternatively the area becomes $A' = Q/w' = 12/0.028 = 428.6\ \text{m}^2$, i.e. about $25\%$ larger.

Practical adoption: Provide a settling length of about 22 m (round up from 21.4 m) with width 20 m and depth 3.0 m, plus additional inlet-transition and outlet lengths (each roughly $0.25$–$0.5$ of the basin length). The 80% factor adds the safety margin needed to ensure the 0.30 mm particles are reliably trapped despite turbulence.

(a) Function and types

A surge tank is a free-water-surface storage placed between the long low-pressure conduit (headrace tunnel) and the pressure conduit (penstock). Its functions are: (i) to absorb/relieve water-hammer pressure rises by reflecting the pressure wave so the long tunnel is protected; (ii) to supply water to the turbine during sudden load increase before the tunnel flow accelerates, and to store water during load rejection; (iii) to improve the speed-regulation (governing) of the plant.

Three types: simple (cylindrical) surge tank, restricted-orifice (throttled) surge tank, and differential (Johnson) surge tank. (Others: gallery/expansion-chamber type.)

(b) Maximum upsurge and period (frictionless)

Velocity in tunnel: $v_0 = Q/A_t = 18/9.0 = 2.0\ \text{m/s}.$

For a frictionless simple surge tank, conservation of energy between the surging mass in the tunnel and the rise in the tank gives the maximum surge amplitude:

$$Z_{max} = v_0 \sqrt{\frac{L A_t}{g A_s}}.$$

Evaluate the term under the root:

$$\frac{L A_t}{g A_s} = \frac{1800 \times 9.0}{9.81 \times 45} = \frac{16\,200}{441.45} = 36.70\ \text{s}^2.$$ $$\sqrt{36.70} = 6.058\ \text{s}.$$ $$Z_{max} = 2.0 \times 6.058 = 12.12\ \text{m} \approx \textbf{12.1 m above reservoir level}.$$

Period of the (undamped) oscillation:

$$T = 2\pi \sqrt{\frac{L A_s}{g A_t}}.$$ $$\frac{L A_s}{g A_t} = \frac{1800 \times 45}{9.81 \times 9.0} = \frac{81\,000}{88.29} = 917.4\ \text{s}^2,\quad \sqrt{917.4} = 30.29\ \text{s}.$$ $$T = 2\pi \times 30.29 = 190.3\ \text{s} \approx \textbf{190 s} \;(\approx 3.2\ \text{min}).$$

(c) Effect of tunnel friction

In reality friction dissipates energy. For an upsurge (load rejection) friction opposes the rising flow into the tank and therefore reduces the maximum upsurge below the frictionless value of 12.1 m. The oscillation is also damped, decaying over several cycles to the new steady level. Conversely, for the first downsurge the friction loss makes the lowest level deeper than the frictionless estimate. The frictionless $Z_{max}$ is thus a conservative (over-estimate) for the upsurge and is often used for freeboard fixing, while damped analysis is used for accurate design.

(a) Velocity, critical time, sudden-closure surge

Pipe area: $A = \dfrac{\pi D^2}{4} = \dfrac{\pi (1.4)^2}{4} = \dfrac{\pi \times 1.96}{4} = 1.539\ \text{m}^2.$

Velocity: $v = \dfrac{Q}{A} = \dfrac{6.0}{1.539} = 3.898\ \text{m/s} \approx \textbf{3.90 m/s}.$

Critical (pipe reflection) time:

$$T_c = \frac{2L}{a} = \frac{2 \times 400}{1100} = \frac{800}{1100} = 0.727\ \text{s} \approx \textbf{0.73 s}.$$

Joukowsky (instantaneous closure) pressure rise:

$$\Delta H = \frac{a\,v}{g} = \frac{1100 \times 3.898}{9.81} = \frac{4287.8}{9.81} = 437.1\ \text{m} \approx \textbf{437 m of water}.$$

Total head at valve (gross static + surge):

$$H_{total} = 250 + 437.1 = 687.1\ \text{m} \approx \textbf{687 m of water}.$$

This is the worst-case (instantaneous) value and governs the safety check.

(b) Slow closure in 8 s

Since $T_{close} = 8\ \text{s} > T_c = 0.73\ \text{s}$, the closure is slow (gradual) — the reflected wave returns long before the valve is fully shut, so the full Joukowsky surge does not develop. Use the Michaud (Allievi) slow-closure estimate:

$$\Delta H = \frac{2 L v}{g\, T_{close}} = \frac{2 \times 400 \times 3.898}{9.81 \times 8} = \frac{3118.4}{78.48} = 39.7\ \text{m} \approx \textbf{39.7 m of water}.$$

Total head at valve $\approx 250 + 39.7 = 289.7\ \text{m} \approx \textbf{290 m}$ — far below the instantaneous case, confirming the benefit of slow closure.

(c) Reasons for slow closure and a structural consequence

Reasons slow closure is preferred:

1. It drastically reduces the water-hammer pressure rise (here from 437 m to 40 m), so the penstock wall and anchors need not be designed for the extreme Joukowsky head.
2. It avoids damaging pressure spikes/fatigue on valves, joints and the spiral casing, improving safety and life.

Structural consequence of water hammer: the penstock shell thickness must be sized for the maximum internal pressure (static + surge) using the thin-cylinder hoop-stress relation $t = \dfrac{p\,D}{2\,\sigma_{allow}} + c$ (plus corrosion allowance $c$); anchor blocks and expansion joints must also resist the transient thrust. Inadequate provision can cause pipe bursting or buckling.

(a) Specific speed and turbine selection

Metric specific speed (power form, $P$ in kW, $H$ in m):

$$N_s = \frac{N\sqrt{P}}{H^{5/4}}.$$

With $P = 22\,000\ \text{kW}$:

• $\sqrt{P} = \sqrt{22\,000} = 148.32.$
• $H^{5/4} = 95^{1.25}$. Compute: $\ln 95 = 4.5539$; $\times1.25 = 5.6924$; $e^{5.6924} = 296.6$.

$$N_s = \frac{500 \times 148.32}{296.6} = \frac{74\,162}{296.6} = 250.0 \approx \textbf{250 (metric, kW)}.$$

For $H \approx 95\ \text{m}$ and $N_s \approx 250$, the value lies in the Francis turbine range (Francis: roughly $N_s \approx 60$–$400$ metric; heads $\sim 30$–$300\ \text{m}$). A Pelton ($N_s \lesssim 35$ per jet, very high head) is unsuitable at 95 m, and a Kaplan ($N_s \gtrsim 300$–$1000$, low head) is unsuitable at this head. Select a (medium-speed) Francis turbine. ✔

(b) Number of poles

Synchronous-speed relation: $N = \dfrac{120 f}{p}\Rightarrow p = \dfrac{120 f}{N}.$

$$p = \frac{120 \times 50}{500} = \frac{6000}{500} = 12\ \text{poles} \;(\textbf{12 poles}, i.e. 6 pole-pairs).$$

This is an even integer, so the direct coupling at 500 rpm is feasible. ✔

(c) Impulse vs reaction turbines and the draft tube

Why a draft tube is essential for reaction turbines: A reaction runner is set above the tailwater and the water leaves it with appreciable pressure and velocity. The gradually diverging draft tube (i) recovers the residual kinetic energy at the runner exit by converting it to pressure (reducing exit losses, raising efficiency), and (ii) allows the runner to be set above tailwater while still using the full head down to the tailrace by creating a partial vacuum (suction head). For an impulse turbine the water already leaves the buckets at atmospheric pressure with little useful kinetic energy and the runner must sit above the tailwater discharging freely to atmosphere; a draft tube would serve no purpose (and any submergence would drown the buckets), so it is not used.

(a) Classification of hydropower plants

By head:

• Low head (typically $< 30\ \text{m}$): large discharge, Kaplan/propeller turbines, often river-bed plants.
• Medium head ($\approx 30$–$300\ \text{m}$): Francis turbines common.
• High head ($> 300\ \text{m}$): small discharge, long penstocks, Pelton/Turgo turbines.

By flow availability / operation:

• Storage (reservoir) plant: a dam stores monsoon water in a reservoir and releases it on demand; can firm up flow and provide seasonal regulation and peaking.
• Run-of-river (RoR) plant: little or no storage; generation follows the natural river flow, so output is high in the wet season and low in the dry season.
• Peaking run-of-river (PRoR): has a small pondage (hours to a day) to re-time daily flow and meet peak demand while still essentially passing the river flow.
• Pumped-storage plant: pumps water to an upper reservoir during off-peak (cheap) hours and generates during peaks; a net energy consumer but valuable for peaking/storage of energy.

(b) Nepalese context

Most Nepalese schemes are run-of-river because: large storage dams require huge reservoirs, large investment, long lead times, resettlement and submergence of land in steep valleys, and carry seismic and sediment (rapid reservoir-sedimentation) risks; RoR plants are cheaper, quicker to build and well-suited to steep rivers with high heads. The drawback is large seasonal variation (dry-season power deficit), which is why peaking/storage projects are increasingly pursued.

Examples: Kulekhani I and II (Nepal's principal storage/peaking reservoir scheme) provide dry-season peaking. A prominent reservoir/storage scheme under development is the Budhi Gandaki (and the proposed Nalsing Gad / Tanahu Seti storage projects). Large RoR examples include Upper Tamakoshi (456 MW, PRoR).

(a) Forebay vs surge tank

• A forebay is an enlarged basin (a small pond) at the end of an open/free-surface headrace canal, just before the penstock intake. Because the headrace is already at atmospheric pressure, the forebay simply provides a small pondage and a smooth transition into the penstock, and traps any remaining floating debris/sediment. It is provided when the headrace is an open canal.
• A surge tank is provided when the conveyance is a closed pressure conduit (tunnel/pipe); it gives a free water surface to relieve water-hammer pressures and supply/store water during load changes. It is provided for long pressurised headrace tunnels.

In short: forebay ↔ open-channel headrace (no pressure-transient role); surge tank ↔ pressurised tunnel (controls water hammer).

(b) Required forebay plan area

Required pondage volume = discharge × time:

$$V = Q \times t = 10\ \text{m}^3/\text{s} \times (15 \times 60)\ \text{s} = 10 \times 900 = 9000\ \text{m}^3.$$

Plan area from usable depth fluctuation $\Delta h = 2.5\ \text{m}$:

$$A = \frac{V}{\Delta h} = \frac{9000}{2.5} = 3600\ \text{m}^2 \approx \textbf{3600 m}^2.$$

(E.g. a basin roughly $90\ \text{m} \times 40\ \text{m}$.)

(c) Three essential components/features of a forebay

1. A trash rack at the penstock intake to stop floating debris.
2. A spillway / overflow weir to discharge excess water safely during load rejection.
3. A gate / control & flushing arrangement (penstock intake gate plus a bottom flushing/scour outlet for settled sediment); often with sufficient submergence over the penstock mouth to prevent air-entraining vortices.

(a) Annual energy generated

Rated annual energy $= P_{inst}\times 8760\ \text{h}$, actual energy $=$ rated $\times$ capacity factor:

$$E = 30\,000\ \text{kW} \times 8760\ \text{h} \times 0.55.$$

Step: $30\,000 \times 8760 = 262\,800\,000\ \text{kWh} = 262.8\ \text{GWh (rated)}.$

$$E = 262.8 \times 0.55 = 144.54\ \text{GWh} \approx \textbf{144.5 GWh/yr} \;(= 1.4454\times10^8\ \text{kWh}).$$

(b) Annual cost and unit generation cost

Annual fixed charges $= 0.11 \times 4.5\times10^9 = \text{NRs } 495\times10^6 = \text{NRs } 495\ \text{million}.$

Total annual cost $=$ fixed charges $+$ O&M:

$$C_{annual} = 495 + 60 = \text{NRs } 555\ \text{million} = 5.55\times10^8\ \text{NRs/yr}.$$

Unit generation cost:

$$c = \frac{C_{annual}}{E} = \frac{5.55\times10^8\ \text{NRs}}{1.4454\times10^8\ \text{kWh}} = 3.840\ \text{NRs/kWh} \approx \textbf{NRs 3.84 per kWh}.$$

Step check: $5.55/1.4454 = 3.840.$

(c) Load factor vs capacity factor

• Load factor $= \dfrac{\text{average load}}{\text{peak (maximum) load over the period}}$. It describes how uniformly the load/demand is spread; it is a property of the load curve.
• Capacity (plant) factor $= \dfrac{\text{actual energy generated}}{\text{energy if run at installed capacity for the whole period}} = \dfrac{\text{average load}}{\text{installed capacity}}.$ It describes how fully the plant is utilised.

Difference: they share the same numerator (average load) but different denominators — load factor uses the maximum demand, capacity factor uses the installed capacity. If maximum demand equals installed capacity the two are equal; since installed capacity $\geq$ peak load generally, capacity factor $\leq$ load factor.

(a) Cavitation and Thoma's factor

Cavitation is the formation of vapour bubbles in regions where the local pressure in the flowing water falls to (or below) the vapour pressure — typically on the suction side of the runner blades and at the draft-tube inlet. When these bubbles are carried to higher-pressure zones they collapse violently, producing pitting/erosion of the metal, noise, vibration and loss of efficiency.

Thoma's cavitation factor is the dimensionless number

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

where $H_s$ is the suction head (runner set above tailwater). Cavitation is avoided as long as the actual $\sigma$ is greater than the critical value $\sigma_c$ for that runner; $\sigma_c$ rises with specific speed, so high-specific-speed runners must be set lower (smaller or even negative $H_s$).

(b) Maximum permissible suction head

At the limiting condition $\sigma = \sigma_c$:

$$\sigma_c = \frac{H_a - H_v - H_s}{H} \;\Rightarrow\; H_s = H_a - H_v - \sigma_c H.$$

Substitute:

$$H_s = 9.6 - 0.3 - (0.10 \times 80) = 9.6 - 0.3 - 8.0 = 1.3\ \text{m}.$$ $$\boxed{H_s \le \textbf{1.3 m above tailwater}.}$$

The runner must be set no higher than 1.3 m above the tailwater level; a lower setting (more submergence) gives an additional safety margin.

(c) Two measures to reduce cavitation damage

1. Set the runner lower (reduce or make negative the suction head) so that $\sigma > \sigma_c$ with a margin.
2. Use cavitation-resistant materials / surfacing (e.g. stainless-steel runners or weld-overlay on vulnerable zones) and maintain smooth, well-profiled blade surfaces; admitting a small amount of air to the draft tube also cushions bubble collapse.

(a) Surface vs underground powerhouse

• Surface powerhouse: built at ground level (usually near the tailrace/river). Advantage: lower construction cost, simpler access, ventilation and crane handling, easier construction and maintenance.
• Underground powerhouse: excavated as a cavern inside the rock mass. Advantage: protection from rockfall/avalanche/flood, shorter pressure conduits in steep terrain, security, and freedom from surface land/topographic constraints (good for high-head Himalayan sites).

(b) Components of a surface powerhouse (described layout)

            ____________________________________
           |  Superstructure: roof, walls,      |
           |  EOT gantry/travelling crane,      |
           |  control room, switchgear          |
           |------------------------------------|
  Penstock |  Intermediate (generator floor):   |
  =====>===|  generator, governor, valves,      |
           |  turbine inlet / spiral casing     |
           |------------------------------------|
           |  Substructure: turbine, draft tube,|
           |  draft-tube gates -> TAILRACE ===> |
           |____________________________________|

• Substructure (below generator floor): foundation, scroll/spiral casing, turbine runner, draft tube and draft-tube gates discharging to the tailrace. It transmits loads to the foundation and houses the water-conducting parts.
• Intermediate part (generator/machine floor): turbine main shaft, generator, governor, main inlet valve, auxiliaries.
• Superstructure: building enclosure (walls, roof), the EOT travelling crane for erection/maintenance, control room, switchgear and ventilation.

(c) Two factors governing choice of an underground powerhouse

1. Geology/rock quality — sound, self-supporting rock is needed to form a stable cavern economically.
2. Topography & hazards — steep, narrow gorges or high-head sites where there is no safe surface bench, or where avalanche/landslide/flood protection (and security) makes a cavern preferable.

(a) Jet velocity and jet diameter

Jet velocity from the nozzle:

$$V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 320}.$$

$2 \times 9.81 \times 320 = 6278.4$; $\sqrt{6278.4} = 79.24\ \text{m/s}$.

$$V_1 = 0.98 \times 79.24 = 77.66\ \text{m/s} \approx \textbf{77.7 m/s}.$$

Jet diameter from continuity, $Q = \dfrac{\pi}{4}d^2 V_1$:

$$d = \sqrt{\frac{4Q}{\pi V_1}} = \sqrt{\frac{4 \times 0.85}{\pi \times 77.66}} = \sqrt{\frac{3.40}{243.97}} = \sqrt{0.013936} = 0.1181\ \text{m}.$$ $$d \approx \textbf{0.118 m} \;(118\ \text{mm}).$$

(b) Wheel (pitch) diameter and jet ratio

Bucket (peripheral) velocity:

$$u = \phi\sqrt{2gH} = 0.46 \times 79.24 = 36.45\ \text{m/s}.$$

Wheel diameter from $u = \dfrac{\pi D N}{60}$:

$$D = \frac{60\,u}{\pi N} = \frac{60 \times 36.45}{\pi \times 600} = \frac{2187.0}{1884.96} = 1.160\ \text{m} \approx \textbf{1.16 m}.$$

Jet ratio:

$$m = \frac{D}{d} = \frac{1.160}{0.1181} = 9.82 \approx \textbf{9.8}.$$

The usual acceptable range is $m \approx 10$–$14$ (sometimes 6–35). A value of $\approx 9.8$ is at the low but acceptable end for a single-jet wheel; if a larger ratio were wanted, a lower speed (larger $D$) or smaller jet (more jets) could be used.

(c) Water power, shaft power, number of buckets

Water (available) power:

$$P_{water} = \rho g Q H = 9.81 \times 0.85 \times 320 = 2668.3\ \text{kW} \approx \textbf{2.67 MW}.$$

Step: $9.81\times0.85 = 8.3385$; $\times320 = 2668.3\ \text{kW}$.

Shaft (output) power using overall efficiency:

$$P_{shaft} = \eta_o\, P_{water} = 0.85 \times 2668.3 = 2268.1\ \text{kW} \approx \textbf{2.27 MW}.$$

(The hydraulic efficiency $\eta_h = 0.88$ represents the runner/wheel conversion; the runner (wheel) power would be $0.88 \times 2668.3 = 2348\ \text{kW}$, and the remaining mechanical/volumetric losses bring the shaft output down to $\eta_o\,P_{water} = 2268\ \text{kW}$.)

Number of buckets (Taygun's rule, $z = 0.5\,m + 15$):

$$z = 0.5 \times 9.82 + 15 = 4.91 + 15 = 19.9 \approx \textbf{20 buckets}.$$

Providing about 20 buckets ensures no jet water escapes between successive buckets, which is satisfactory for this wheel.