BE Civil Engineering (IOE, TU) Hydropower Engineering (IOE, CE 755) Question Paper 2078 Nepal

(a) Definitions

• Gross head ($H_g$): the vertical difference in water-surface elevation between the intake/forebay and the tailrace, measured under static (no-flow) conditions.
• Net head ($H_n$): the head actually available to the turbine after deducting all hydraulic losses (trashrack, intake, headrace, penstock friction and bend losses): $H_n = H_g - \sum h_L$.
• Firm power: the power that can be guaranteed for (almost) the entire period, i.e. the power corresponding to the dependable minimum flow (often the 95–100 % exceedance discharge).

(b) Installed (design) capacity

Design discharge = discharge available 40 % of the time $= Q_{40} = 26\ \text{m}^3/\text{s}$.

$$P = \eta\,\gamma\,Q\,H = 0.82 \times 9.81 \times 26 \times 95$$ $$P = 0.82 \times 9.81 = 8.0442;\quad 8.0442 \times 26 = 209.15;\quad 209.15 \times 95 = 19{,}869\ \text{kW}$$

Installed capacity ≈ 19.87 MW (≈ 19.9 MW).

(c) Mean annual energy

The turbine passes $Q_{turb} = \min(Q,\ 26)$ and shuts off when $Q < 7$. Compute power at each exceedance level using $P = \eta\gamma Q_{turb}H$ with $\eta\gamma H = 0.82\times9.81\times95 = 764.18$ kW per (m³/s).

Use the average-of-ordinates (mid-point of each 10 % band) to get mean power. Average successive pairs and weight each band by 10 % (0.1):

• 0–10: avg of (P at 10) only edge → use trapezoidal between listed points. Pair averages (kW): (19869+19869)/2=19869; 19869; 19869; (19869+16812)/2=18341; (16812+14519)/2=15666; (14519+12227)/2=13373; (12227+9934)/2=11081; (9934+7642)/2=8788; (7642+5349)/2=6496.

These 9 band-averages cover the 10 %→100 % range (each band = 10 % = 0.1). Add half-band contributions at the two ends (assume P constant beyond 10 % at 19,869 kW for the 0–10 band, and equal to 5,349 kW for the 100 % tail which is the shutdown limit).

Mean power:

$$\bar P = 0.1\,[19869 + (19869+19869+18341+15666+13373+11081+8788+6496)]$$

Sum inside: $19869+19869+19869+18341+15666+13373+11081+8788+6496 = 133352$ kW; including the leading 0–10 band (19,869):

$$\bar P = 0.1 \times (19869 + 133352)/...$$

Simpler trapezoidal over the 10 listed ordinates (9 intervals of 0.1 each):

$$\bar P = 0.1\Big[\tfrac{19869+5349}{2} + (19869+19869+19869+16812+14519+12227+9934+7642)\Big]$$

Interior sum $=19869+19869+19869+16812+14519+12227+9934+7642 = 120741$ kW. End terms $=(19869+5349)/2 = 12609$ kW.

$$\bar P = 0.1\,(120741 + 12609) = 0.1 \times 133350 = 13{,}335\ \text{kW} \approx 13.34\ \text{MW}$$

This covers 90 % of the year (10 %→100 % exceedance). For the top 10 % (0–10 %) the plant is capped at 19,869 kW. Add that band: contribution to the year-average $= 0.1 \times 19869 = 1986.9$ kW, while the 13.34 MW above already represents the 0.9 fraction:

$$\bar P_{year} = 0.9 \times 13335 + 0.1 \times 19869 = 12001.5 + 1986.9 = 13{,}988\ \text{kW} \approx 13.99\ \text{MW}$$

Annual energy (8760 h):

$$E = 13988\ \text{kW} \times 8760\ \text{h} = 1.2254\times10^{8}\ \text{kWh}$$

Mean annual energy ≈ 122.5 GWh/year, giving a plant (capacity) factor $= 13.99/19.87 = 0.70$.

(a) Need for settling basins (high-head plants)

High-head plants use Pelton/Francis turbines running at high velocities; sediment (especially hard quartz $\ge 0.2$ mm) acts as an abrasive and rapidly erodes nozzles, needles, runner buckets and guide vanes, reducing efficiency and life. A settling basin reduces water velocity so suspended particles settle out before entering the headrace/penstock.

Four design requirements:

1. Trap the target particle size (e.g. $\ge 0.2$–$0.25$ mm) with the required efficiency.
2. Low enough through-flow velocity to allow settling (non-scouring of deposited silt).
3. Smooth inlet/transition to avoid turbulence and short-circuiting; uniform flow distribution.
4. Provision for flushing/removal of trapped sediment (continuous or intermittent), and adequate freeboard.

(b) Basin dimensions (ideal theory)

Cross-sectional area required for the through-flow velocity:

$$A = \frac{Q}{v} = \frac{12}{0.30} = 40\ \text{m}^2$$

Choose a practical depth-to-width relationship. Adopt depth $D = 4$ m (typical for medium plants). Then

$$B = \frac{A}{D} = \frac{40}{4} = 10\ \text{m}$$

Ideal length: a particle entering at the surface must reach the floor (fall depth $D$) before it is carried out:

$$L = \frac{v\,D}{w} = \frac{0.30 \times 4}{0.025} = 48\ \text{m}$$

Ideal basin: $B = 10$ m, $D = 4$ m, $L = 48$ m. (Surface-loading check: $Q/(BL)=12/(10\times48)=0.025$ m/s $= w$ ✓.)

(c) Turbulence (Camp) correction

Corrected fall velocity:

$$w' = w - 0.04\sqrt{v} = 0.025 - 0.04\sqrt{0.30} = 0.025 - 0.04(0.5477) = 0.025 - 0.02191 = 0.00309\ \text{m/s}$$

Corrected length:

$$L' = \frac{v\,D}{w'} = \frac{0.30 \times 4}{0.00309} = \frac{1.20}{0.00309} = 388\ \text{m}$$

Comment: The turbulence correction increases the required length almost 8-fold (48 m → ≈ 388 m), which is impractical for a single chamber. The very large factor shows that the empirical constant 0.04 makes the corrected settling velocity nearly collapse; in practice a milder turbulence factor (e.g. $L' = \alpha L$ with $\alpha \approx 1.2$–$1.5$) or multiple narrower parallel chambers with lower velocity ($v \approx 0.2$ m/s) are adopted to keep length reasonable while still trapping 0.25 mm particles.

(a) Wave celerity

$$a = \frac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{K}{E}\dfrac{D}{t}}}$$

Numerator: $\sqrt{K/\rho} = \sqrt{2.1\times10^{9}/1000} = \sqrt{2.1\times10^{6}} = 1449.1$ m/s.

Denominator factor: $\dfrac{K}{E}\dfrac{D}{t} = \dfrac{2.1\times10^{9}}{200\times10^{9}} \times \dfrac{1.6}{0.014} = 0.0105 \times 114.29 = 1.200$.

$$a = \frac{1449.1}{\sqrt{1+1.200}} = \frac{1449.1}{\sqrt{2.200}} = \frac{1449.1}{1.4832} = 977.0\ \text{m/s}$$

$a \approx 977$ m/s.

(b) Sudden closure (Joukowsky)

$$\Delta H = \frac{a\,V_0}{g} = \frac{977.0 \times 4.0}{9.81} = \frac{3908}{9.81} = 398.4\ \text{m}$$

Total maximum head at valve:

$$H_{max} = H_0 + \Delta H = 180 + 398.4 = 578.4\ \text{m}$$

$\Delta H \approx 398$ m; $H_{max} \approx 578$ m.

(c) Closure classification for $T = 6$ s

Critical (reflection) time:

$$T_c = \frac{2L}{a} = \frac{2 \times 420}{977.0} = \frac{840}{977.0} = 0.86\ \text{s}$$

Since $T = 6\ \text{s} > T_c = 0.86\ \text{s}$, this is a slow closure. Use Michaud/Allievi slow-closure formula:

$$\Delta H_{slow} = \frac{2 L V_0}{g\,T} = \frac{2 \times 420 \times 4.0}{9.81 \times 6} = \frac{3360}{58.86} = 57.1\ \text{m}$$

Total head $\approx 180 + 57.1 = 237.1$ m — a large reduction compared with sudden closure (578 m → 237 m).

Role of surge tank: A surge tank placed near the powerhouse end of a long low-pressure conduit converts the conduit into a short penstock; it reflects/absorbs the pressure surge (mass oscillation) so that the rapid water-hammer pressure is confined to the short penstock below the tank, protecting the long headrace tunnel and reducing the effective $L$ in the water-hammer equation.

(a) Specific speed and turbine type

Metric power specific speed (per unit, single runner):

$$N_s = \frac{N\sqrt{P}}{H_n^{5/4}}, \quad P\ \text{in kW}$$ $$H_n^{5/4} = 310^{1.25};\ \ln310 = 5.7366;\times1.25 = 7.1708;\ e^{7.1708} = 1301$$ $$N_s = \frac{500\sqrt{24000}}{1301} = \frac{500 \times 154.92}{1301} = \frac{77460}{1301} = 59.5$$

$N_s \approx 60$ (metric, kW). For a single jet this falls in the Pelton range ($N_s \approx 10$–$60$ for multi-jet up to ~70). Combined with the high head (310 m), a Pelton turbine is recommended (impulse turbine; suits high head, moderate discharge).

(b) Pelton jet and runner sizing

Discharge required (assume $\eta_o = 0.88$):

$$Q = \frac{P}{\eta_o\,\gamma\,H_n} = \frac{24{,}000}{0.88 \times 9.81 \times 310} = \frac{24000}{2676.2} = 8.97\ \text{m}^3/\text{s}$$

Jet velocity:

$$V_1 = C_v\sqrt{2gH_n} = 0.98\sqrt{2 \times 9.81 \times 310} = 0.98\sqrt{6082.2} = 0.98 \times 77.99 = 76.4\ \text{m/s}$$

Bucket (peripheral) speed:

$$u = \phi V_1 = 0.46 \times 76.4 = 35.16\ \text{m/s}$$

Runner pitch diameter from $u = \pi D N/60$:

$$D = \frac{60\,u}{\pi N} = \frac{60 \times 35.16}{\pi \times 500} = \frac{2109.6}{1570.8} = 1.343\ \text{m}$$

$D \approx 1.34$ m.

Total jet area: $A = Q/V_1 = 8.97/76.4 = 0.1174\ \text{m}^2$.

Max allowed single-jet diameter $d_{max} = D/12 = 1.343/12 = 0.1119$ m; single-jet area $= \frac{\pi}{4}(0.1119)^2 = 0.00984\ \text{m}^2$.

Number of jets:

$$n = \frac{A}{a_{jet}} = \frac{0.1174}{0.00984} = 11.9 \Rightarrow \textbf{12 jets (use 6 nozzles on a vertical-shaft twin runner, or 2 wheels of 6 jets)}.$$

Since a single Pelton wheel is limited to ~6 jets, two wheels (6 jets each) on one shaft would be adopted; revised $N_s$ per runner then drops accordingly. (Acceptable answer: 12 jets total satisfying $d \le D/12$.)

(c) Pelton governing

        penstock
           |
        [ spear/needle ] --> moves in/out of nozzle (controls jet area)
           ||  jet
           VV
        ===O=== runner (buckets)
        deflector plate (swings into jet on sudden load drop)

When load on the generator falls, the governor (oil servomotor driven by a centrifugal/electronic speed sensor) must reduce the jet discharge to avoid overspeed. To prevent water hammer the spear (needle) is moved slowly to reduce the nozzle opening. For sudden large load rejection, a deflector is first swung into the jet to instantly divert water away from the buckets (no sudden flow change in the penstock, so no water hammer); the spear then closes gradually. On load increase the spear opens to admit more flow. This two-element (deflector + spear) action keeps the runner speed (and hence frequency) constant while limiting pressure rise.

(a) Net saleable energy

Gross generation:

$$E_{gross} = P \times CF \times 8760 = 30{,}000\ \text{kW} \times 0.60 \times 8760\ \text{h} = 157.68\times10^{6}\ \text{kWh} = 157.68\ \text{GWh}$$

After 4 % losses:

$$E_{net} = 157.68 \times (1 - 0.04) = 157.68 \times 0.96 = 151.4\ \text{GWh}$$

Net saleable energy ≈ 151.4 GWh/year.

(b) Annual revenue requirement

Loan = $0.70 \times 5400 = 3780$ M NPR; Equity = $0.30 \times 5400 = 1620$ M NPR.

Debt service (capital recovery over 12 yr at 10 %): CRF $= \dfrac{i(1+i)^n}{(1+i)^n-1}$ with $i=0.10,\ n=12$.

$(1.10)^{12} = 3.1384$. CRF $= \dfrac{0.10 \times 3.1384}{3.1384 - 1} = \dfrac{0.31384}{2.1384} = 0.14676$.

Annual debt service $= 3780 \times 0.14676 = 554.8$ M NPR.

Return on equity (14 % of equity): $= 0.14 \times 1620 = 226.8$ M NPR.

O&M (1.5 % of installed cost): $= 0.015 \times 5400 = 81.0$ M NPR.

Annual revenue requirement $= 554.8 + 226.8 + 81.0 = \textbf{862.6 M NPR/year}$.

(c) Required average tariff

$$\text{Tariff} = \frac{\text{Annual revenue requirement}}{\text{Net energy}} = \frac{862.6\times10^{6}\ \text{NPR}}{151.4\times10^{6}\ \text{kWh}} = 5.70\ \text{NPR/kWh}$$

Required tariff ≈ NPR 5.7/kWh (during the 12-year loan-repayment phase; after the loan is retired the debt-service component disappears and the tariff falls to roughly $(226.8+81.0)/151.4 \approx$ NPR 2.0/kWh).

Comment: A levelised price near NPR 5–6/kWh during repayment is broadly in line with NEA's wet-/dry-season PPA rates for RoR projects (historically around NPR 4.8 wet / NPR 8.4 dry). The project is competitive provided generation is firm; the high front-loaded debt burden means tariffs are highest early and drop sharply after year 12, which is typical of Nepali hydropower economics. Sensitivity to capacity factor (dry-season flow) and interest rate is the main risk.

(i) Classification by head

(ii) Classification by operation / load

• Base-load plant: runs continuously near full capacity (high plant factor); typically large storage/RoR with firm flow.
• Peak-load plant: operates during peak-demand hours only (low plant factor, high capacity); needs storage or pondage.
• Pumped-storage plant: pumps water to an upper reservoir off-peak and generates during peak.

Run-of-river vs Storage plant

For a frictionless simple surge tank, sudden complete load rejection produces undamped mass oscillation.

Maximum surge amplitude (frictionless):

$$Z_{max} = V_0\sqrt{\frac{L\,A_t}{g\,A_s}}$$ $$\frac{L A_t}{g A_s} = \frac{1500 \times 9}{9.81 \times 30} = \frac{13500}{294.3} = 45.87\ \text{s}^2$$ $$Z_{max} = 2.2 \times \sqrt{45.87} = 2.2 \times 6.773 = 14.90\ \text{m}$$

$Z_{max} \approx 14.9$ m above reservoir level.

Period of oscillation:

$$T = 2\pi\sqrt{\frac{L\,A_s}{g\,A_t}}$$ $$\frac{L A_s}{g A_t} = \frac{1500 \times 30}{9.81 \times 9} = \frac{45000}{88.29} = 509.7\ \text{s}^2$$ $$T = 2\pi\sqrt{509.7} = 2\pi \times 22.58 = 141.8\ \text{s}$$

$T \approx 142$ s (≈ 2.4 min). Friction would reduce the amplitude and slightly alter the period in practice.

Forebay vs Surge tank

Functions of a forebay

1. Provides temporary storage (pondage) to meet sudden load demand and absorb load fluctuations.
2. Allows final settling of fine sediment and removal of floating debris (trashrack).
3. Maintains the required submergence/head over the penstock intake to prevent air entrainment (vortex).
4. Provides a smooth transition from open-channel to pressure flow and a place for spillway/overflow during load rejection.

Functions of an intake

1. Divert/admit the required design discharge into the conveyance system with minimum head loss.
2. Control flow (gates) and exclude floating debris, ice and bed-load sediment (trashrack, sill, gravel trap).
3. Prevent entry of air (adequate submergence, anti-vortex design).
4. Permit shut-off for maintenance and protect downstream structures during floods.

Velocity

$$A = \frac{\pi}{4}D^2 = \frac{\pi}{4}(1.2)^2 = 1.131\ \text{m}^2,\quad V = \frac{Q}{A} = \frac{6.0}{1.131} = 5.31\ \text{m/s}$$

Friction head loss (Darcy–Weisbach)

$$h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.018 \times \frac{250}{1.2} \times \frac{(5.31)^2}{2\times9.81}$$ $$= 0.018 \times 208.33 \times \frac{28.20}{19.62} = 3.75 \times 1.4373 = 5.39\ \text{m}$$

$h_f \approx 5.39$ m.

Power lost in friction

$$P_{loss} = \gamma Q h_f = 9.81 \times 6.0 \times 5.39 = 317.3\ \text{kW}$$

$P_{loss} \approx 317$ kW.

Economic diameter principle: A larger diameter reduces velocity and hence friction loss (more energy/revenue) but increases the capital cost of the pipe; a smaller diameter is cheaper but wastes energy. The economic diameter is the one that minimises the total annual cost = (annualised capital cost of penstock) + (annual cost of lost energy). It is found where $\dfrac{d}{dD}(\text{total annual cost}) = 0$, i.e. where the marginal saving in energy cost equals the marginal increase in capital cost.

Thoma's cavitation factor (plant/available value)

$$\sigma = \frac{H_{atm} - H_v - H_s}{H} = \frac{10.3 - 0.24 - 3.5}{60} = \frac{6.56}{60} = 0.1093$$

Safety check: Cavitation is avoided when $\sigma \ge \sigma_c$.

$$\sigma = 0.1093 > \sigma_c = 0.095 \quad\Rightarrow\ \textbf{SAFE against cavitation.}$$

Margin: $0.1093 - 0.095 = 0.0143$. The setting has a small but positive margin. (Maximum permissible suction head for the critical condition: $H_{s,max} = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.095\times60 = 10.06 - 5.70 = 4.36$ m; since actual $H_s = 3.5\ \text{m} < 4.36\ \text{m}$, the setting is acceptable.)

Function of the draft tube

1. Recovers a large part of the kinetic energy at the runner exit by gradual expansion (converts velocity head to pressure head), increasing the effective/net head and efficiency.
2. Allows the turbine to be set above the tailwater while still utilising the head between the runner exit and tailrace (creates negative/suction pressure at outlet).
3. By controlling the outlet pressure it helps keep pressure above vapour pressure, mitigating cavitation when correctly designed and submerged.

(a) Components of a surface powerhouse

• Sub-structure: the foundation and the water-passage components below the generator floor — spiral/scroll casing, draft tube, turbine pit, and the concrete substructure supporting the machines.
• Super-structure: the building above the generator floor housing the generator, control room, switchgear, overhead travelling (EOT) crane for erection/maintenance, ventilation and auxiliaries. Typically divided into machine hall, erection bay, and service/control bays.

(b) Hydropower potential of Nepal

• Theoretical potential: ≈ 83,000 MW (about 83 GW), based on Dr. Hari Man Shrestha's 1966 estimate.
• Technically feasible: ≈ 45,000 MW.
• Economically feasible: ≈ 42,000 MW.
• Status (≈2078 BS): installed capacity has grown to the order of ~2,000+ MW (NEA + IPPs combined), still only a small fraction (~2–3 %) of the economically feasible potential; mostly RoR, with Kulekhani as the only seasonal-storage plant. Large projects (Upper Tamakoshi 456 MW) commissioned around this period.

(c) Pondage and pondage factor

• Pondage is the small storage (typically for a few hours to a day) provided in/just upstream of a RoR plant (e.g. in the forebay or a daily-regulating pond) to meet short-duration peak demand that exceeds the instantaneous river flow.
• Pondage factor $= \dfrac{\text{maximum (peak) discharge demanded}}{\text{average daily river flow}}$. A pondage factor > 1 means the plant can peak above the run-of-river flow using stored water, recharged during off-peak hours.

(d) Advantages of underground powerhouse (Himalayan terrain)

1. Suits steep, narrow, unstable valleys where no flat surface site exists.
2. Protected from landslides, rockfall, avalanches and floods (GLOFs).
3. Shorter/steeper penstock or pressure shaft, often reducing conveyance length.
4. Lower environmental/visual impact and better security; rock provides natural support and thermal stability. (Disadvantage: higher excavation cost and need for good rock and careful geotechnical investigation.)