BE Civil Engineering (IOE, TU) Hydropower Engineering (IOE, CE 755) Question Paper 2076 Nepal

(a) Categories of hydropower potential (Nepal figures)

• Theoretical potential – the gross power available from all flowing water assuming 100% conversion and full exploitation of every drop and head. For Nepal this is about 83,000 MW (~83 GW).
• Technical potential – the portion that can be developed using current engineering technology, accounting for site access, geology and head/flow limits. For Nepal this is about 42,000–45,000 MW.
• Economically feasible potential – the part of the technical potential that can be developed at a cost competitive with alternatives. For Nepal this is about 42,000 MW in the classic studies (often quoted alongside the technical figure).

(b) Installed capacity

$$P = \rho g Q H \eta_o$$ $$P = 1000 \times 9.81 \times 12 \times 85 \times 0.86$$ $$P = 8{,}604{,}900\ \text{W} \approx 8.605\ \text{MW}$$

Installed capacity $P \approx 8.61\ \text{MW}$.

(c) Annual energy

With capacity factor $C_f = 0.55$ and $8760$ hours/year:

$$E = P \times C_f \times 8760$$ $$E = 8.605\ \text{MW} \times 0.55 \times 8760\ \text{h} = 41{,}460\ \text{MWh}$$

Annual energy $E \approx 41.5\ \text{GWh}$.

(d) A run-of-river plant has little or no storage, so its output tracks the natural river hydrograph. In the monsoon (Jun–Sep) snowmelt plus rainfall keep discharge well above the design flow, so the plant runs near full capacity. In the dry season (Dec–Apr) base flow can fall to a small fraction of the design discharge, so the turbines must be throttled or shut off, sharply reducing the capacity factor. This seasonal energy deficit is why Nepal historically imported power in winter and why storage/PRoR projects are emphasised.

(a) Four functional zones (longitudinal flow shown left to right):

  inlet                                              outlet
   |   transition   |        settling zone        |  weir |
  ==>~~~~~~~~~~~~~~~~|=============================|~~~~~~~==> to headrace
   |                |   ....particles settle....  |       |
   |________________|_____________________________|_______|
            \                                         /
             \----------- SEDIMENT (storage) --------/  --> flushing gate

1. Inlet / transition zone – spreads the concentrated intake flow gently across the full basin width so velocity is uniform and low; abrupt expansion is avoided to prevent turbulence and short-circuiting.
2. Settling (working) zone – the main length where the horizontal velocity is low enough that particles settle to the bed before reaching the outlet.
3. Sediment storage zone – the hopper/sloping floor below the settling zone where deposited sediment collects between (or during) flushes.
4. Outlet zone – an overflow weir/transition that draws clarified water uniformly into the headrace without re-suspending deposited material.

(b) Sizing

Settling length (a particle entering at the surface must reach the bed before leaving):

$$\frac{L}{v_h} = \frac{d}{w} \;\Rightarrow\; L = \frac{v_h\, d}{w} = \frac{0.30 \times 1.5}{0.021} = 21.43\ \text{m}$$

Width from continuity ($Q = v_h\, d\, W$):

$$W = \frac{Q}{v_h\, d} = \frac{8.0}{0.30 \times 1.5} = 17.78\ \text{m}$$

Overflow-rate (surface-loading) check – ideal-basin theory requires the surface area $A_s = Q/w$:

$$A_{s,\text{req}} = \frac{Q}{w} = \frac{8.0}{0.021} = 380.95\ \text{m}^2$$ $$A_{s,\text{provided}} = L \times W = 21.43 \times 17.78 = 381.0\ \text{m}^2$$

The provided plan area equals the required surface-loading area, confirming the design is consistent. Adopt $L \approx 21.5\ \text{m}$, $W \approx 17.8\ \text{m}$, $d = 1.5\ \text{m}$ (a settling efficiency factor of 1.2–1.5 is normally applied to $L$ for turbulence; e.g. $L \approx 26$–$32\ \text{m}$ in practice).

(c) Continuous (hopper-bottom, e.g. Bieri/Serpent-Sand or Dufour) flushing keeps removing sediment while the plant runs, so the basin never silts up during the flood season when Himalayan rivers carry extreme suspended loads (often >5000 ppm). Intermittent flushing requires periodic shutdown and wastes water and generation; with very high inflow concentrations the basin would fill faster than it could be emptied, allowing abrasive sediment to pass to the turbines and accelerate runner erosion. Hence continuous flushing protects the runners and maximises availability.

(a) Economic-diameter factors. The optimum diameter minimises the sum of annualised capital cost (increases with $D$, since more steel/excavation) and annual energy-loss cost (decreases with $D$, because friction loss $\propto 1/D^5$ for a given $Q$). Governing factors: discharge $Q$, available head, length, steel/fabrication cost, energy value (tariff), plant load/capacity factor, interest rate and design life, and a velocity limit (usually 4–6 m/s to control water hammer and loss).

cost
 |  \ capital cost                /
 |   \                          /  total cost
 |    \____                ____/
 |         \___        ___/   <- minimum = economic D
 |             \______/
 |     energy-loss cost (falling)
 +---------------------------------- diameter D -->

(b) Velocity, friction loss, net head

Area: $A = \dfrac{\pi}{4}D^2 = \dfrac{\pi}{4}(1.8)^2 = 2.545\ \text{m}^2$

Velocity: $V = \dfrac{Q}{A} = \dfrac{12}{2.545} = 4.716\ \text{m/s}$ (within 4–6 m/s — acceptable)

Friction head loss (Darcy–Weisbach):

$$h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.018 \times \frac{420}{1.8} \times \frac{4.716^2}{2 \times 9.81}$$ $$h_f = 0.018 \times 233.33 \times 1.1336 = 4.76\ \text{m}$$

Net head:

$$H_n = H_g - h_f = 180 - 4.76 = 175.24\ \text{m}$$

Net head $H_n \approx 175.2\ \text{m}$.

(c) Power output

Turbine (shaft) power:

$$P_t = \rho g Q H_n \eta_t = 1000 \times 9.81 \times 12 \times 175.24 \times 0.88$$ $$P_t = 18{,}154{,}000\ \text{W} \approx 18.15\ \text{MW}$$

Electrical power at generator terminals:

$$P_e = P_t \times \eta_g = 18.154 \times 0.96 = 17.43\ \text{MW}$$

Turbine power $\approx 18.15\ \text{MW}$; electrical power $\approx 17.4\ \text{MW}$.

(a) Water hammer is the transient pressure surge produced when the flow velocity in a closed conduit is changed rapidly (e.g. valve/guide-vane closure). The kinetic energy of the moving water column is converted into a pressure (elastic) wave that travels back and forth at celerity $a$. The critical (or phase) time is the time for the pressure wave to travel to the reservoir and return:

$$T_{cr} = \frac{2L}{a}$$

• Rapid closure: $T_c \le T_{cr}$ → the reflected relief wave has not yet returned, so the full Joukowsky head develops.
• Slow closure: $T_c > T_{cr}$ → relief waves reduce the peak, and the surge is given by the slow-closure (Michaud/Allievi) formula.

(b) Instantaneous closure (Joukowsky) and critical time

$$\Delta H_{max} = \frac{a\,V}{g} = \frac{1100 \times 4.716}{9.81} = 528.8\ \text{m}$$ $$T_{cr} = \frac{2L}{a} = \frac{2 \times 420}{1100} = 0.764\ \text{s}$$

Since $T_c = 6\ \text{s} > T_{cr} = 0.764\ \text{s}$, the closure is slow and the Joukowsky value is not used for design.

(c) Slow-closure surge, design head, thickness

Michaud (Allievi) slow-closure surge:

$$\Delta H = \frac{2 L V}{g\, T_c} = \frac{2 \times 420 \times 4.716}{9.81 \times 6} = 67.3\ \text{m}$$

Design head (static gross + surge):

$$H_d = H_g + \Delta H = 180 + 67.3 = 247.3\ \text{m}$$

Design internal pressure:

$$p = \rho g H_d = 1000 \times 9.81 \times 247.3 = 2.426 \times 10^6\ \text{Pa} = 2.426\ \text{MPa}$$

Required shell thickness (thin-cylinder hoop formula with joint efficiency):

$$t = \frac{p\,D}{2\,\sigma\,\eta_j} = \frac{2.426\times10^6 \times 1.8}{2 \times 110\times10^6 \times 0.90} = 0.02205\ \text{m} = 22.05\ \text{mm}$$

Adding a corrosion allowance of ~1.5 mm gives a fabricated thickness of about 24 mm.

Surge head $\approx 67.3\ \text{m}$, design head $\approx 247.3\ \text{m}$, shell thickness $\approx 22\ \text{mm}$ (≈24 mm with corrosion allowance).

(a) Definitions

• Firm (primary) power – the power that can be guaranteed to be available essentially at all times (corresponds to the minimum dependable flow/head), used to meet base demand.
• Secondary power – the surplus power available only when flow exceeds the firm value (e.g. in the monsoon); it is non-guaranteed and usually sold cheaply or dumped.
• Load factor – ratio of average load to peak load over a period: $LF = \dfrac{\text{average load}}{\text{peak load}}$.
• Plant (capacity) factor – ratio of actual energy generated to the energy that would be generated at full installed capacity over the same period: $C_f = \dfrac{E_{actual}}{P_{installed}\times 8760}$.

(b) Annual costs

Capital recovery factor (CRF):

$$\text{CRF} = \frac{i(1+i)^n}{(1+i)^n - 1} = \frac{0.10(1.10)^{30}}{(1.10)^{30}-1}$$

With $(1.10)^{30} = 17.449$:

$$\text{CRF} = \frac{0.10 \times 17.449}{17.449 - 1} = \frac{1.7449}{16.449} = 0.10608$$

Annualised capital cost:

$$A_{cap} = C \times \text{CRF} = 2.4\times10^9 \times 0.10608 = \text{NRs}\ 254.6\ \text{million}$$

Annual O&M:

$$A_{om} = 0.025 \times 2.4\times10^9 = \text{NRs}\ 60.0\ \text{million}$$

Total annual cost:

$$A_{total} = 254.6 + 60.0 = \text{NRs}\ 314.6\ \text{million}$$

(c) Unit cost and BCR

Levelised generation cost:

$$c = \frac{A_{total}}{E} = \frac{314.6\times10^6}{42\times10^6} = \text{NRs}\ 7.49/\text{kWh}$$

Annual revenue (benefit):

$$B = E \times \text{tariff} = 42\times10^6 \times 8.50 = \text{NRs}\ 357.0\ \text{million}$$

Benefit–cost ratio:

$$\text{BCR} = \frac{B}{A_{total}} = \frac{357.0}{314.6} = 1.135$$

Comment: The levelised cost (NRs 7.49/kWh) is below the tariff (NRs 8.50/kWh) and $\text{BCR} = 1.135 > 1$, so the project is financially viable, with about a 13.5% margin of benefits over costs. The margin is modest, so the result is sensitive to discount rate, generation shortfall and tariff — a sensitivity analysis is advisable.

(i) Classification by head

(Boundaries vary slightly between textbooks; some take low <15 m, high >250 m.)

(ii) Classification by operating regime / storage

• Run-of-river (RoR): no storage; output follows river flow (most Nepali plants).
• Peaking RoR (PRoR): small pondage for a few hours of daily peaking (e.g. Upper Tamakoshi has limited pondage).
• Storage (reservoir) plants: large reservoir for seasonal regulation and firm dry-season power (e.g. proposed Budhi Gandaki, existing Kulekhani — Nepal's only major storage scheme).
• Pumped-storage plants: pump water to an upper reservoir off-peak and generate at peak (under study in Nepal).

Geometry

Area: $A = (b + zy)y = (2.5 + 1\times1.2)\times1.2 = 3.7 \times 1.2 = 4.44\ \text{m}^2$

Wetted perimeter: $P = b + 2y\sqrt{1+z^2} = 2.5 + 2(1.2)\sqrt{2} = 2.5 + 3.394 = 5.894\ \text{m}$

Hydraulic radius: $R = \dfrac{A}{P} = \dfrac{4.44}{5.894} = 0.7533\ \text{m}$

Manning velocity

$$V = \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.018}(0.7533)^{2/3}(0.0008)^{1/2}$$ $$V = 55.556 \times 0.8276 \times 0.02828 = 1.30\ \text{m/s}$$

Discharge

$$Q = A V = 4.44 \times 1.30 = 5.78\ \text{m}^3/\text{s}$$

Check: $V = 1.30\ \text{m/s}$ lies in the usual non-silting / non-scouring band for a lined canal (about 0.6–3 m/s). It is high enough to keep fine sediment in suspension (no silting) yet well below the scour limit for a concrete lining, so the velocity is acceptable.

Result: $A = 4.44\ \text{m}^2$, $R = 0.753\ \text{m}$, $V = 1.30\ \text{m/s}$, $Q \approx 5.78\ \text{m}^3/\text{s}$.

(a) Functions of a surge tank

• Provides a free water surface close to the turbines so that water-hammer pressure waves in the long tunnel are reflected/relieved and do not travel up the headrace tunnel.
• Supplies/absorbs the differential flow during load changes (gives water on sudden load increase, stores it on load rejection), improving turbine governing/regulation.
• Stores the rising water (upsurge) on load rejection and dampens the mass oscillation of the tunnel water column.

Difference from a forebay: a forebay sits between an open headrace (canal) and the penstock, acting mainly as a small balancing reservoir/transition with a spillway; a surge tank is used with a closed pressure tunnel and is specifically designed to handle pressure transients (water hammer) and mass oscillation. A surge tank must be tall enough to contain the surge; a forebay relieves to atmosphere through a spillway.

(b) Maximum upsurge (frictionless)

For a simple surge tank with frictionless tunnel, the amplitude of mass oscillation is:

$$Z_{max} = V_0\sqrt{\frac{L\,A_t}{g\,A_s}}$$ $$Z_{max} = 2.2\sqrt{\frac{2500 \times 7.0}{9.81 \times 28}}$$ $$Z_{max} = 2.2\sqrt{\frac{17500}{274.68}} = 2.2\sqrt{63.71} = 2.2 \times 7.982 = 17.56\ \text{m}$$

Maximum upsurge $Z_{max} \approx 17.6\ \text{m}$ above reservoir level. (Friction reduces the actual upsurge; the tank height above static level is fixed at this value plus freeboard.)

(a) Specific speed

Power-form specific speed (metric, $P$ in kW, $H$ in m):

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{600\sqrt{8000}}{300^{1.25}}$$

$\sqrt{8000} = 89.44$; $300^{1.25} = 300 \times 300^{0.25} = 300 \times 4.162 = 1248.6$

$$N_s = \frac{600 \times 89.44}{1248.6} = \frac{53{,}666}{1248.6} = 42.98$$

$N_s \approx 43$. This value (roughly 8–30 per jet for Pelton, up to ~70 for multi-jet) together with the high head (300 m) points to a Pelton (impulse) turbine, likely a 2- to 4-jet machine. (A single-jet Pelton has $N_s \approx 10$–$30$; $N_s = 43$ implies a multi-jet Pelton, since $N_s$ per jet $= 43/\sqrt{n_{jets}}$.)

(b) Justification and comparison

• At $H = 300\ \text{m}$ a Francis runner would suffer high pressures and cavitation risk near the outlet; an impulse (Pelton) turbine handles high head efficiently and is favoured above ~250–300 m.
• Pelton vs Francis:
  1. Head range / sediment: Pelton suits high head and tolerates abrasive Himalayan sediment better (buckets/needles are easier to repair); Francis suits medium head but its runner erodes badly under high sediment.
  2. Part-load efficiency: Pelton holds high efficiency over a wide flow range (jets can be shut individually), whereas Francis efficiency drops more sharply at part load.

Selection: a multi-jet Pelton turbine ($N_s \approx 43$).

(a) Powerhouse components (super-, intermediate- and sub-structure):

• Main units: turbine(s), generator(s) on a common vertical shaft, governor and main inlet valve (MIV).
• Water passages: spiral (scroll) casing, guide-vane/wicket-gate mechanism, runner, and draft tube leading to the tailrace.
• Electrical/mechanical auxiliaries: excitation system, busbars, step-up transformers, switchgear/control room, cooling water and lubrication systems, drainage/dewatering pumps, compressed-air system, and the EOT crane for erection/maintenance.
• Structure: machine hall (superstructure), generator floor, turbine floor and the substructure (draft-tube and foundation).

(b) Cavitation, draft tube, Thoma factor

• Cavitation: where local pressure in the flowing water falls below the vapour pressure (typically at the runner outlet / draft-tube inlet of a reaction turbine), vapour bubbles form and then collapse violently in higher-pressure zones, causing pitting, vibration, noise and efficiency loss.
• Draft tube role: a gradually expanding (diverging) conduit from runner exit to tailwater that (i) recovers part of the exit kinetic energy as a pressure rise (regains head), and (ii) allows the runner to be set above tailwater while still using the full head — but its setting must avoid sub-vapour pressures.
• Thoma's cavitation factor:

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

where $H_a$ = atmospheric pressure head, $H_v$ = vapour-pressure head, $H_s$ = suction head (runner setting above tailwater), $H$ = net head. To avoid cavitation, $\sigma \ge \sigma_c$ (the critical value, which depends on specific speed); a higher specific-speed runner requires a larger $\sigma_c$ and hence a lower (or submerged) setting.

(i) Major projects

• Upper Tamakoshi (456 MW) – Nepal's largest plant (commissioned 2021), a peaking-RoR scheme on the Tamakoshi, developed largely with domestic financing.
• Kali Gandaki A (144 MW) and Kulekhani (92 MW, the only seasonal-storage plant) are other landmarks; large storage projects such as Budhi Gandaki (~1200 MW) are under development to provide dry-season firm power.

(ii) Technical / environmental challenges

• Sediment / Himalayan geology: very high suspended-sediment loads cause severe turbine and waterway abrasion; young, fragile geology brings landslides, GLOF risk and difficult tunnelling.
• Hydrological seasonality and climate risk: large monsoon-vs-dry-season flow variation lowers dry-season output, and glacier-fed flows are changing with climate; environmental/e-flow release requirements reduce usable flow.

(iii) Institutional / policy issues

• Transmission, evacuation and market: weak transmission grid and limited domestic demand cause spillage in the monsoon; cross-border trade (power-export agreements with India/Bangladesh) is being developed but constrained by grid and policy.
• Financing, licensing and land/social issues: lengthy licensing (PPA, survey/generation licences), financing-cost and currency risk, plus land acquisition, resettlement and benefit-sharing with local communities slow project delivery.

Overall Nepal has shifted from chronic load-shedding to seasonal surplus, but firm (storage) capacity, grid evacuation, sediment management and export markets remain the key issues.