BE Civil Engineering (IOE, TU) Hydropower Engineering (IOE, CE 755) Question Paper 2079 Nepal

(a) Net head and installed power

Head loss $h_L = 0.06 \times 185 = 11.1\ \text{m}$.

Net head:

$$H_n = H_g - h_L = 185 - 11.1 = 173.9\ \text{m}$$

Installed power at $Q = 12.0\ \text{m}^3/\text{s}$:

$$P = \rho\, g\, Q\, H_n\, \eta = 1000 \times 9.81 \times 12.0 \times 173.9 \times 0.84$$ $$P = 1000 \times 9.81 \times 12.0 = 117720;\quad \times 173.9 = 20{,}471{,}508;\quad \times 0.84 = 17{,}196{,}067\ \text{W}$$

Installed power $\approx 17.20\ \text{MW}$.

(b) Mean annual energy

Power at each flow block (same $H_n$, $\eta$, factor $k = \rho g H_n \eta = 9.81\times173.9\times0.84\times1000 = 1{,}433{,}006\ \text{W per (m}^3/\text{s})$):

Block energies:

• Block 1: $17.196\ \text{MW} \times 2628\ \text{h} = 45{,}191\ \text{MWh} = 45.19\ \text{GWh}$
• Block 2: $12.897\ \text{MW} \times 3504\ \text{h} = 45{,}191\ \text{MWh} = 45.19\ \text{GWh}$
• Block 3: $7.165\ \text{MW} \times 2628\ \text{h} = 18{,}830\ \text{MWh} = 18.83\ \text{GWh}$

Total annual energy:

$$E = 45.19 + 45.19 + 18.83 = 109.21\ \text{GWh}$$

Mean annual energy $\approx 109.2\ \text{GWh}$.

(c) Plant (capacity) factor

Maximum possible energy at installed capacity:

$$E_{max} = 17.196\ \text{MW} \times 8760\ \text{h} = 150{,}637\ \text{MWh} = 150.64\ \text{GWh}$$ $$\text{Capacity factor} = \frac{109.21}{150.64} = 0.725$$

Plant capacity factor $\approx 0.73$ (72.5%).

Comment: A capacity factor of about 73% is high and typical of a run-of-river (RoR) plant on a river with a relatively steady flow regime. RoR plants have little or no storage, so they cannot shift generation to peak hours; they generate whenever water is available and therefore tend to run at high capacity factors but offer limited peaking/firm-power flexibility compared with reservoir (storage) plants.

(a) Settling basin functions and zones

A settling basin reduces the flow velocity so that suspended sediment settles out before water enters the headrace/penstock. Hard quartz sand carried by Himalayan rivers abrades turbine runners, guide vanes and seals; removing particles above a chosen size (commonly $0.2\ \text{mm}$ for medium-head Pelton/Francis units) protects the plant and reduces maintenance.

   inlet transition      settling (working) zone        outlet
  (flow expands,        (low velocity, particles        (weir / sill,
   velocity drops)       settle to hopper)               clear water out)
  ====\                                                   /====
       \____________________________________________   __/
        :  :   . .  .   .  .  .   .  .  .  .  . . .  : :  -> headrace
        :  :__________ sediment collects in hopper __: :
        flushing gate / sluice at bottom for periodic flushing

Three zones:

• Inlet (transition) zone: gradually expands the cross-section so velocity drops uniformly without jets or eddies that would re-suspend sediment.
• Settling (working) zone: long, wide, shallow-velocity region where particles fall to a collection hopper; flushing sluices remove deposits.
• Outlet zone: an overflow weir / sill that draws clear surface water into the headrace while leaving deposited sediment behind.

(b) Plan dimensions

Ideal settling theory: a particle entering at the surface must reach the bottom before leaving the basin. The governing relation is the surface-loading (overflow-rate) criterion:

$$\frac{Q}{A_{surface}} = w \quad\Rightarrow\quad A_{surface} = \frac{Q}{w}$$

Required surface area:

$$A_{surface} = \frac{9.0}{0.021} = 428.6\ \text{m}^2$$

Basin width from continuity (cross-section $= B \times D$ carries $Q$ at velocity $v$):

$$B = \frac{Q}{v\,D} = \frac{9.0}{0.30 \times 3.0} = \frac{9.0}{0.90} = 10.0\ \text{m}$$

Settling length from surface area:

$$L = \frac{A_{surface}}{B} = \frac{428.6}{10.0} = 42.9\ \text{m}$$

Cross-check by the fall-time / horizontal-travel method: time to settle $t = D/w = 3.0/0.021 = 142.9\ \text{s}$; horizontal travel $L = v\,t = 0.30 \times 142.9 = 42.9\ \text{m}$ (consistent).

Required surface area $\approx 429\ \text{m}^2$, width $B = 10.0\ \text{m}$, ideal length $L \approx 42.9\ \text{m}$. In practice a turbulence/safety factor of about 1.2-1.5 is applied, giving an adopted length of roughly $52$-$64\ \text{m}$.

(a) Rapid vs slow closure and pressure rise

Critical (round-trip) time of the pressure wave:

$$T_c = \frac{2L}{a} = \frac{2 \times 520}{1100} = \frac{1040}{1100} = 0.945\ \text{s}$$

Since the closure time $T = 4.0\ \text{s} > T_c = 0.945\ \text{s}$, the closure is slow. Use the Michaud (slow-closure) formula.

Flow velocity:

$$A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(2.0)^2 = 3.1416\ \text{m}^2,\qquad V = \frac{Q}{A} = \frac{9.0}{3.1416} = 2.865\ \text{m/s}$$

Michaud head rise:

$$\Delta H = \frac{2 L V}{g\,T} = \frac{2 \times 520 \times 2.865}{9.81 \times 4.0} = \frac{2979.6}{39.24} = 75.9\ \text{m}$$

Water-hammer (surge) head rise $\Delta H \approx 75.9\ \text{m}$. (For comparison, full Joukowsky rapid closure would give $aV/g = 1100\times2.865/9.81 = 321.3\ \text{m}$, which the slow closure avoids.)

(b) Design surge head and wall thickness

Maximum head the penstock must withstand:

$$H_{max} = H + \Delta H = 173.9 + 75.9 = 249.8\ \text{m}$$

Internal pressure:

$$p = \rho g H_{max} = 1000 \times 9.81 \times 249.8 = 2{,}450{,}538\ \text{Pa} = 2.451\ \text{MPa}$$

Thin-cylinder hoop stress with joint efficiency $\eta_j$:

$$\sigma = \frac{p\,d}{2\,t\,\eta_j} \;\Rightarrow\; t = \frac{p\,d}{2\,\sigma_{allow}\,\eta_j}$$ $$t = \frac{2.451\times10^6 \times 2.0}{2 \times 120\times10^6 \times 0.90} = \frac{4.902\times10^6}{2.16\times10^8} = 0.02269\ \text{m}$$ $$t = 22.7\ \text{mm}$$

Adding a corrosion allowance of about $1.5\ \text{mm}$ gives an adopted plate thickness of roughly $24\ \text{mm}$.

Design surge head $H_{max} \approx 249.8\ \text{m}$; required wall thickness $t \approx 22.7\ \text{mm}$ (adopt $\approx 24\ \text{mm}$).

(a) Specific speed and turbine selection

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{600\sqrt{16000}}{173.9^{1.25}}$$

Numerator: $\sqrt{16000} = 126.49$, so $600 \times 126.49 = 75{,}895$.

Denominator: $173.9^{1.25}$. $\ln 173.9 = 5.1585$; $\times 1.25 = 6.4481$; $e^{6.4481} = 631.9$.

$$N_s = \frac{75{,}895}{631.9} = 120.1$$

Specific speed $N_s \approx 120$ (metric, kW).

Selection: With $H = 173.9\ \text{m}$ (medium-high head) and $N_s \approx 120$, the operating point falls squarely in the Francis turbine range. Typical metric specific-speed bands are: Pelton $\approx 10$-$35$ (single jet) up to $\sim 70$ (multi-jet), Francis $\approx 60$-$400$, Kaplan/propeller $\approx 300$-$1000$. A head near $175\ \text{m}$ with $N_s \approx 120$ is well suited to a Francis runner, which gives good efficiency over this medium-head, moderate-flow regime.

(b) Effect of large discharge / low head, and operating ranges

If the discharge were very large at low head, the same power could only be developed with much higher flow per unit head, which pushes the specific speed up. High $N_s$ (above ~300) and low head (below ~50 m) call for an axial-flow Kaplan (or propeller) turbine, whose adjustable blades pass large flows efficiently. Conversely, very high head with small flow lowers $N_s$ and favours an impulse Pelton wheel.

Text sketch of operating ranges:

  Head (m)
  ^
 1500 | Pelton (high head, low Ns ~10-70)
      |  \
  500 |   \____ Francis (medium head, Ns ~60-400)
      |        \
  150 |         \______
      |                \___ Kaplan/Propeller (low head, high Ns ~300-1000)
   10 |                      \___________________
      +-------------------------------------------> Specific speed Ns

General rule: as head decreases and flow increases, specific speed rises and the preferred turbine shifts Pelton -> Francis -> Kaplan.

(a) Forebay vs surge tank

• A forebay is an enlarged open basin at the end of a (low-pressure, free-surface) headrace canal, just upstream of the penstock intake. It acts as a small balancing reservoir that supplies water for short-duration demand fluctuations and traps any remaining sediment/trash before the penstock. It is used in canal (open-channel) conveyance schemes.
• A surge tank is a vertical shaft/chamber on a long pressurized headrace tunnel, placed as close as practicable to the powerhouse. Its purpose is to absorb water-hammer/pressure transients and supply or store water during load changes, protecting the tunnel and reducing penstock surge. It is used in tunnel (pressurized) conveyance schemes.

(b) Surge amplitude and period (frictionless simple surge tank)

For a frictionless simple surge tank, the maximum surge amplitude is:

$$Z_{max} = V_0\sqrt{\frac{L\,A_t}{g\,A_s}}$$

Substituting:

$$\frac{L\,A_t}{g\,A_s} = \frac{2400 \times 9.0}{9.81 \times 60} = \frac{21600}{588.6} = 36.70\ \text{s}^2$$ $$\sqrt{36.70} = 6.058\ \text{s}$$ $$Z_{max} = 1.0 \times 6.058 = 6.06\ \text{m}$$

Maximum surge amplitude $Z_{max} \approx 6.06\ \text{m}$ above reservoir level.

Period of (undamped) oscillation:

$$T = 2\pi\sqrt{\frac{L\,A_s}{g\,A_t}} = 2\pi\sqrt{\frac{2400 \times 60}{9.81 \times 9.0}}$$ $$\frac{2400 \times 60}{9.81 \times 9.0} = \frac{144000}{88.29} = 1631.0\ \text{s}^2,\quad \sqrt{1631.0} = 40.39\ \text{s}$$ $$T = 2\pi \times 40.39 = 253.8\ \text{s}$$

Period $T \approx 253.8\ \text{s}$ (about $4.2$ minutes).

(i) Classification by head:

• Low head: $H < 30\ \text{m}$ — e.g. run-of-river barrage plants on large lowland rivers (Kaplan units).
• Medium head: $30\ \text{m} \leq H \leq 300\ \text{m}$ — e.g. Francis-turbine plants on hill rivers.
• High head: $H > 300\ \text{m}$ — e.g. Pelton-turbine plants fed by long penstocks in steep terrain.

(Exact band limits vary slightly between texts; the medium band is sometimes quoted as 25-250 m.)

(ii) Classification by operation / storage:

• Run-of-river (RoR): little/no storage; generates with available flow (e.g. most Nepali medium-head plants).
• Storage (reservoir) plant: a reservoir stores seasonal flow to provide firm/peaking power (e.g. a multi-purpose dam scheme).
• Pumped-storage plant: pumps water to an upper reservoir during off-peak hours and generates during peak demand, acting as energy storage.

Summary: head governs turbine choice; storage governs how flexibly the plant can meet demand.

Major components of a surface powerhouse:

Structural / civil:

• Substructure (draft-tube and turbine foundations, scroll-case block)
• Intermediate structure (generator floor, control gallery)
• Superstructure (machine hall, service bay, crane gantry, roof)

Electro-mechanical:

• Turbines and governors
• Generators (alternators) with excitation systems
• Main inlet valves / butterfly or spherical valves
• Step-up transformers and switchgear/busbars
• Overhead travelling (EOT) crane
• Cooling, dewatering, drainage and auxiliary systems
• Control room / SCADA and protection equipment

Functions of three components:

1. Turbine: converts the hydraulic energy (head and flow) of water into mechanical shaft power.
2. Generator: converts the turbine's mechanical rotation into electrical energy.
3. Main inlet valve: isolates the turbine from the penstock for maintenance and provides emergency shut-off, protecting the unit.
4. EOT crane: lifts and positions heavy runners, rotors and valves during erection and maintenance.

Geometry of a trapezoidal section (side slope $m = 1$, so each side adds $m\,y$ horizontally):

Area:

$$A = (b + m y)\,y = (3.0 + 1\times1.5)\times1.5 = 4.5 \times 1.5 = 6.75\ \text{m}^2$$

Wetted perimeter:

$$P = b + 2y\sqrt{1+m^2} = 3.0 + 2(1.5)\sqrt{2} = 3.0 + 3.0\times1.4142 = 3.0 + 4.2426 = 7.243\ \text{m}$$

Hydraulic radius:

$$R = \frac{A}{P} = \frac{6.75}{7.243} = 0.9320\ \text{m}$$

Manning's equation:

$$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$

$R^{2/3}$: $\ln 0.9320 = -0.07041$; $\times \tfrac{2}{3} = -0.04694$; $e^{-0.04694} = 0.9542$.

$S^{1/2} = \sqrt{0.0005} = 0.022361$.

$$Q = \frac{1}{0.015}\times 6.75 \times 0.9542 \times 0.022361$$ $$= 66.667 \times 6.75 \times 0.9542 \times 0.022361$$ $$66.667 \times 6.75 = 450.0;\quad \times 0.9542 = 429.4;\quad \times 0.022361 = 9.60\ \text{m}^3/\text{s}$$

Computed capacity $Q \approx 9.60\ \text{m}^3/\text{s} > 9.0\ \text{m}^3/\text{s}$ design discharge.

The section at $y = 1.5\ \text{m}$ carries about $9.6\ \text{m}^3/\text{s}$, which exceeds the required $9.0\ \text{m}^3/\text{s}$ by roughly $7\%$, so the depth is adequate and provides a small margin (a freeboard of $0.3$-$0.5\ \text{m}$ should still be added above the water surface).

(a) Capital recovery factor and annual capital charge

Capital recovery factor:

$$CRF = \frac{i(1+i)^n}{(1+i)^n - 1},\quad i = 0.10,\ n = 25$$

Compute $(1.10)^{25}$: $\ln 1.10 = 0.0953102$; $\times 25 = 2.382755$; $e^{2.382755} = 10.8347$.

$$CRF = \frac{0.10 \times 10.8347}{10.8347 - 1} = \frac{1.08347}{9.8347} = 0.11017$$

$CRF \approx 0.1102$.

Annual capital (debt-service) charge:

$$= CRF \times \text{capital} = 0.11017 \times 5.16\times10^9 = 5.685\times10^8\ \text{NRs/yr}$$

Annual capital charge $\approx \text{NRs } 568.5\ \text{million/yr}$.

(b) Levelised generation cost

Annual O&M cost:

$$= 0.02 \times 5.16\times10^9 = 1.032\times10^8\ \text{NRs/yr} = \text{NRs } 103.2\ \text{million/yr}$$

Total annual cost:

$$= 568.5 + 103.2 = 671.7\ \text{million NRs/yr} = 6.717\times10^8\ \text{NRs/yr}$$

Annual energy:

$$109.2\ \text{GWh} = 109.2\times10^6\ \text{kWh}$$

Levelised cost:

$$= \frac{6.717\times10^8}{109.2\times10^6} = 6.151\ \text{NRs/kWh}$$

Levelised generation cost $\approx \text{NRs } 6.15\ \text{per kWh}$.

(a) Jet velocity

$$V = C_v\sqrt{2 g H} = 0.98\sqrt{2 \times 9.81 \times 320}$$ $$2\times9.81\times320 = 6278.4;\quad \sqrt{6278.4} = 79.24\ \text{m/s}$$ $$V = 0.98 \times 79.24 = 77.66\ \text{m/s}$$

Jet velocity $V \approx 77.7\ \text{m/s}$.

(b) Jet discharge Jet area:

$$a = \frac{\pi}{4}d^2 = \frac{\pi}{4}(0.100)^2 = 7.854\times10^{-3}\ \text{m}^2$$ $$Q = a V = 7.854\times10^{-3} \times 77.66 = 0.6100\ \text{m}^3/\text{s}$$

Jet discharge $Q \approx 0.610\ \text{m}^3/\text{s}$.

(c) Theoretical jet (kinetic) power

$$P = \tfrac{1}{2}\rho Q V^2 = 0.5 \times 1000 \times 0.6100 \times (77.66)^2$$ $$(77.66)^2 = 6031.1;\quad 0.5\times1000\times0.6100 = 305.0;\quad \times 6031.1 = 1{,}839{,}500\ \text{W}$$

Theoretical jet power $P \approx 1.84\ \text{MW}$.

(Equivalently $P = \rho g Q H \times C_v^2 = 1000\times9.81\times0.610\times320\times0.9604 = 1.84\ \text{MW}$, confirming the result.)

(a) Hydropower potential of Nepal (commonly cited figures)

• Theoretical potential: about $83{,}000\ \text{MW}$ (roughly $83\ \text{GW}$), based on the gross runoff of the country's major river systems (Koshi, Gandaki, Karnali, Mahakali and southern rivers).
• Technically feasible potential: about $42{,}000\ \text{MW}$ (~$42\ \text{GW}$).
• Economically feasible potential: about $42{,}000\ \text{MW}$ is often quoted as technically feasible, with the economically feasible portion commonly cited at roughly the same order; many references state about $42\ \text{GW}$ technically and a large fraction of that economically exploitable under current conditions.

Most developed and planned plants in Nepal are run-of-river (medium head, Francis or Pelton), with only a few storage schemes, leading to a marked seasonal generation deficit in the dry months.

(b) Three major challenges

1. Seasonal flow variability: monsoon-dominated hydrology gives high wet-season flows and very low dry-season flows; with mostly RoR plants and little storage, the country faces winter energy deficits and surplus in the wet season.
2. Difficult terrain, geology and sediment: steep, fragile Himalayan terrain, landslides, glacial-lake outburst flood (GLOF) risk and very high sediment loads raise construction cost and cause turbine abrasion.
3. Financing, transmission and market constraints: high up-front capital cost, limited domestic financing, weak/insufficient transmission and cross-border evacuation infrastructure, and the need for power-trade agreements (e.g. export to India/Bangladesh) to absorb wet-season surplus.

(Other valid challenges: geopolitical/transboundary water issues, right-of-way and resettlement, and institutional/regulatory delays.)