Components of Hydropower System

Economic diameter

The economic diameter of a penstock is the diameter for which the total annual cost (annual cost of the penstock + annual value of energy lost in friction) is minimum.

• A larger diameter gives lower velocity and lower head loss (more energy), but costs more steel, excavation, supports and transport.
• A smaller diameter is cheaper to build, but head loss $h_f = \dfrac{8fLQ^2}{\pi^2 g D^5}$ is high, so energy is lost every year.

Cost
 |T                        T
 | T                     T
 |  T                  T
 |   T     min       T
 |     T T  *  T T        C
 |E         :        C
 | E        :    C
 |   E      : C
 |     E   C:
 |       C  E
 |   C      :  E   E   E
 +----------:-------------> D
         D_econ
 T = total, C = penstock cost,
 E = value of energy lost

Method of determination (analytical/graphical)

1. Fix data: design discharge Q, gross head H, penstock length L, friction factor f, energy rate, interest rate, life, steel price.
2. Assume trial diameters $D_1, D_2, \dots$ (e.g. velocity 2–6 m/s).
3. Penstock cost: for each D, find shell thickness $t = \dfrac{pD}{2\sigma\eta_j}$ + corrosion allowance (p includes water hammer), steel weight $= \pi D t L\rho_s$, then capital cost (steel + fabrication + supports + excavation). Annual cost = capital × capital recovery factor + O&M.
4. Energy loss cost: $h_f = \dfrac{8fLQ^2}{\pi^2 g D^5}$; power lost $= 9.81Qh_f\eta$ kW; annual energy lost $=$ power × hours of operation (× plant factor); its value = energy × tariff.
5. Total annual cost = penstock annual cost + value of lost energy.
6. Plot the three curves against D; the D at the lowest total cost is the economic diameter. Analytically, set $\dfrac{d(\text{total cost})}{dD} = 0$.
7. Round up to a standard size and check velocity, water hammer, and that head loss is usually 2–5% of gross head.

Empirical formulas (for preliminary design)

With P = rated power in kW, H = head in m, D in m:

• Sarkaria (1979): $D = 0.62\dfrac{P^{0.35}}{H^{0.65}}$
• USBR: $D = 0.176\left(\dfrac{P}{H}\right)^{0.466}$
• Fahlbusch (1987): $D = 0.52\,H^{-0.17}\left(\dfrac{P}{H}\right)^{0.43}$
• Velocity method: $D = \sqrt{4Q/(\pi V)}$ with economic velocity about 3–5 m/s.

These give a first estimate, which is refined by the cost analysis above.

Why a settling basin is essential in Nepal

A settling basin (desander) is an enlarged chamber after the intake where flow velocity is reduced so that suspended sediment settles and is flushed back to the river. In Nepal it is essential because:

1. Young, fragile Himalaya: rivers carry very high sediment loads, especially in the monsoon (often several thousand ppm).
2. Hard, sharp quartz particles: quartz and feldspar content is high, so turbines (especially high-head Pelton and Francis) suffer severe abrasion/erosion of runners, needles, guide vanes and seals.
3. Most plants are run-of-river: there is no large reservoir to trap sediment.
4. Sediment fills headrace canals and tunnels, reducing capacity and increasing head loss.
5. Wear causes loss of efficiency, frequent shutdowns and costly repairs (e.g. Jhimruk, Khimti, Kaligandaki A).

Typical layout

 PLAN
        inlet          settling zone       outlet
 from  +-----+--------------------------+-----+ to
 intake| /   |  ====   chamber 1   ==== |   \ |headrace
 ----> |<    |--------------------------|    >|---->
       | \   |  ====   chamber 2   ==== |   / |
       +-----+--------------------------+-----+
        transition    |flushing| spill   transition
                      v channel v weir

 SECTION (longitudinal)
  WL ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
   \ inlet                          outlet|
    \___                        ____weir__|
        \______________________/
         bed slope 1:20-1:50 --> flushing
                               gate/pipe

Design criteria

• Particle size to remove: 0.15–0.2 mm for high-head plants (> 100 m) and 0.2–0.5 mm for low/medium heads.
• Trap efficiency: usually ≥ 90% of the design particle size.
• Flow velocity: less than critical (scour) velocity, by Camp: $V = a\sqrt{d}$ cm/s (d in mm, a = 44 for 0.1–1 mm), normally 0.2–0.4 m/s.
• Size: settling length $L = \dfrac{hV}{w}$ (w = fall velocity), increased for turbulence (Vetter/Camp efficiency); $L/B$ about 4–10.
• Inlet and outlet transitions: gradual (about 1:5 to 1:8) to give uniform flow; baffles or flow-straightening walls.
• Number of chambers: at least two, so one can be flushed or repaired while the other works.
• Flushing: continuous or intermittent (hydraulic flushing gates, Serpent/SSSS systems, pipes); bed slope 2–5% towards the flushing outlet.
• Spillway: to pass excess water during floods or load rejection.
• Provision for sediment monitoring and bypass of flood flows.

Desander

A desander (settling basin) is a structure placed after the intake (and gravel trap) of a hydropower scheme, in which the flow is slowed down so that suspended sand particles larger than a design size settle and are flushed out, protecting the headrace and turbines from abrasion.

         +---------------------------+
 from -->| inlet  | settling zone  | outlet |--> headrace
 intake  | trans. |  V = 0.2-0.4   | weir   |
         +--------+----------------+--------+
                    \____ sediment ____/ -> flushing

Design criteria

1. Design particle size: 0.2 mm for medium heads; 0.1–0.15 mm for high-head Pelton plants; up to 0.5 mm for low heads. Smaller size for hard quartz sediment.
2. Trap efficiency: 90% or more of the design particle size should be removed.
3. Flow velocity: horizontal velocity below critical (scouring) velocity, by Camp's formula $V_c = a\sqrt{d}$ (cm/s, d in mm; a = 44 for 0.1 < d < 1 mm). Typical 0.2–0.4 m/s.
4. Fall velocity: from Stokes' law or standard curves (e.g. about 0.021 m/s for 0.2 mm quartz at 20 °C).
5. Dimensions:
   • Surface area $A = BL = Q/w$ for ideal settling.
   • Length $L = \dfrac{hV}{w}$, corrected for turbulence (Camp or Vetter's equation $\eta = 1 - e^{-wA/Q}$).
   • Depth 3–5 m (plus sediment storage depth); $L/B$ about 4–10.
6. Transitions: gradual inlet expansion (1:5 to 1:8) and outlet contraction to give uniform flow with no dead zones; guide walls/baffles if needed.
7. Number of chambers: at least two parallel chambers so the plant runs during flushing or repair.
8. Flushing arrangement: continuous or intermittent hydraulic flushing with gates/pipes; bed slope about 1:20–1:50; enough head to return sediment to the river.
9. Spillway/overflow: to release excess water.
10. Location: stable ground, close to the intake, with space for flushing return.

An intake is the structure at the head of a water conductor system that draws the required discharge from a river or reservoir while keeping out floating debris, ice, and as much sediment as possible.

1. Side (lateral) intake

Opening in the river bank at an angle to the flow, usually upstream of a diversion weir.

• Suitability: most common in Nepalese run-of-river projects on medium and large rivers. Placing it on the outer bank of a bend reduces bedload entry.

   river flow ---->
 =========================
     |         | weir   |
     |  intake |========|
 ====|  gate   |=========
     v
  to gravel trap / canal

2. Frontal intake

Intake facing the flow directly, with the river bed upstream and a sluice for flushing.

• Suitability: where the river carries little bedload, or in canal-type schemes; strong flow towards the intake but more sediment entry.

3. Bottom (drop, Tyrolean or trench) intake

A trench with an inclined trash rack set in the weir crest; water drops through the rack into a gallery.

• Suitability: steep mountain streams with boulders and coarse bedload, small and micro hydro (common in Nepal's hilly streams).

  flow -->  ___rack___
 ~~~~~~~~~/\/\/\/\/\/\~~~~~~
 =========|  trench  |=====> to canal
          |__________|

4. Tower intake

A vertical tower in a reservoir with ports at several levels, connected to a tunnel.

• Suitability: storage dams with large fluctuation of water level; allows drawing water from different depths.

5. Shore/face intake (dam intake)

Opening in the dam face or reservoir bank, with trash rack and gate, leading to the penstock/tunnel.

• Suitability: concrete dams and pondage schemes (e.g. Kulekhani).

6. Siphon intake

Uses a siphon over the dam crest. Suitable for small heads and small discharges.

A hydraulic tunnel is an underground water conductor (headrace, tailrace, diversion or pressure tunnel) used where a surface canal is not feasible because of steep, unstable terrain or to shorten the route. Its layout and design must satisfy geological, hydraulic, structural and economic criteria.

Layout (alignment) criteria

1. Shortest practical route between intake and surge tank/powerhouse, to reduce cost and head loss.
2. Good geology: pass through sound, uniform rock; avoid faults, shear zones, thrusts, weak and squeezing rock, karst and water-bearing zones. Cross weak zones at right angles.
3. Adequate rock cover: for unlined pressure tunnels, the rock weight above must exceed the internal water pressure (Norwegian criterion $C_{RM} = \dfrac{h\gamma_w F}{\gamma_r\cos\beta}$, F ≈ 1.3–1.5).
4. Stable portals and adits: portals in sound rock, away from landslides and flood levels; adits to give more working faces and reduce construction time.
5. Straight alignment with gentle curves (radius ≥ 5 times the diameter) to reduce head loss.
6. Gradient: enough slope for drainage during construction (about 1:200–1:500); free-flow tunnels follow the hydraulic gradient.

Design criteria

1. Type of flow: pressure tunnel (full flow, circular or horseshoe) or free-flow tunnel (D-shaped/horseshoe, flowing 70–80% full with free-board area of about 15–25%).
2. Size: economic diameter found by comparing excavation/lining cost with value of head loss; minimum size about 2–2.5 m for construction access.
3. Velocity: unlined 1.5–2.5 m/s; concrete-lined 2–4 m/s; steel-lined higher. Velocities must avoid erosion but limit head loss.
4. Shape: circular best for internal pressure; horseshoe/inverted-D easier to excavate by drill and blast and good for external rock pressure.
5. Lining and support: shotcrete, rock bolts, steel ribs, concrete or steel lining depending on rock class (Q-system, RMR), internal pressure and leakage control. Smooth lining reduces Manning's n.
6. Head loss: friction and form losses kept within economic limits.
7. Water hammer and surge: pressure tunnels need a surge tank; check minimum pressure so the crown never goes under negative pressure.
8. Sediment and rock traps near the end of unlined tunnels.
9. Drainage, grouting and access for inspection and maintenance.

A forebay is a small pond or tank at the end of a headrace canal (free-flow system), just upstream of the penstock inlet. It connects the open canal to the pressure pipe.

Functions of forebay

1. Transition from free surface flow in the canal to pressure flow in the penstock.
2. Temporary storage: supplies extra water when the load increases suddenly and stores water when load drops, until the canal flow adjusts.
3. Spills excess water safely through a side spillway on load rejection.
4. Final settling of fine sediment and removal of floating debris with a trash rack.
5. Provides enough submergence over the penstock entrance to prevent vortex and air entry.
6. Acts as a small surge tank for the canal system.

Layout

Main components:

• Inlet transition from the canal, widening gradually.
• Forebay tank: volume usually enough for about 2–3 minutes (some use 30–120 s for small schemes) of design flow.
• Trash rack in front of the penstock mouth, inclined about 60–75°.
• Penstock intake with bellmouth, gate and air vent behind it.
• Spillway/overflow weir and spill channel to the river.
• Flushing (scour) gate at the low point for sediment.
• Minimum submergence of penstock top: $S = cV\sqrt{D}$ (Gordon, c ≈ 0.54 symmetric, 0.72 asymmetric approach).

 PLAN
 headrace          forebay tank
 canal   +------------------------------+
 ======> | transition |      trash rack |  penstock
 ======> |  /         |  pond      ||| |==========>
         | /          |            ||| | gate, vent
         +-----+-------------+-------+--+
               |   spillway  | flush gate
               v  to river   v

 SECTION
 canal WL ~~~~~~~~~~~~~~~~~~~~~~~~~~~
 =====\                       |rack|
       \                      | \  |_____
        \_____________________|__\_| penstock
          sediment --> flush       ===========

Why they are provided

Both are provided between the low-pressure conduit and the penstock to deal with sudden changes of load on the turbines.

• Forebay is provided at the end of an open headrace canal. It changes free-surface flow into pressure flow, stores water for load increases, spills water on load rejection, traps the last sediment and debris, and gives submergence to the penstock inlet.
• Surge tank is provided at the end of a pressure tunnel or long pipeline (as close to the powerhouse as possible). When the turbine gates close or open, it gives a free water surface that absorbs the water hammer, so only the short penstock below it is designed for high water hammer. It also supplies water when load increases and stores it when load decreases, and improves speed regulation.

 Canal:  canal ==> [FOREBAY] ==> penstock ==> PH
 Tunnel: tunnel ==> [SURGE TANK] ==> penstock ==> PH
         (pressure)      free surface

Differences

A surge tank is an open vertical shaft or chamber connected to the pressure conduit (headrace tunnel) at its junction with the penstock, as near the powerhouse as possible. It provides a free water surface that absorbs pressure surges caused by sudden load changes.

  reservoir          surge tank
  ~~~~~~~~~        |  ^   |  max upsurge
  |                |  :   |
  | static WL -----|--:---|-------
  |                |  v   |  min downsurge
  |  headrace tunnel (L, A)|
  |========================+
                            \  penstock
                             \
                              \==> turbine

Functions

1. Reduces water hammer pressure: the water hammer wave is reflected at the tank, so only the short penstock below it is designed for high water hammer; the long tunnel is protected.
2. Load rejection: when turbine gates close, the moving water in the tunnel rises into the tank instead of building up pressure.
3. Load acceptance: when gates open, the tank supplies water immediately until the tunnel flow accelerates, preventing a large pressure drop.
4. Improves governing of the turbine (better speed regulation).
5. Acts as a small reservoir and allows the tunnel to be lighter and cheaper.

Types

Simple (cylindrical), restricted orifice (throttled), differential (Johnson), one-sided/gallery or chamber type, inclined, and air-cushion surge chambers.

Design criteria

1. Location: as close as possible to the powerhouse, on sound rock, to reduce penstock length.
2. Maximum upsurge on full load rejection (frictionless estimate):

$$Z_{max} = V\sqrt{\frac{L A}{g A_s}}$$

where V = tunnel velocity, L = tunnel length, A = tunnel area, $A_s$ = tank area. The top of the tank must be above static level + upsurge + freeboard (or an overflow is provided).

3. Minimum downsurge on load acceptance: the bottom of the tank and the tunnel crown must remain below the lowest water level so that air does not enter the penstock.
4. Stability (Thoma criterion): to damp oscillations, area must exceed

$$A_{th} = \frac{L A V^2}{2 g H_0 h_f}$$

($H_0$ = net head, $h_f$ = tunnel head loss); a factor of safety of 1.5–1.8 is used.

5. Period of oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A}}$, should not resonate with the governor.
6. Structural and economic: shaft lining for rock conditions; choose the type (orifice, differential) to reduce size and cost.

A powerhouse houses the turbines, generators, control and auxiliary equipment. Its plan dimensions and height are fixed mainly by the size and number of generating units.

Parameters for fixing the dimensions

1. Number, type and size of units: turbine type (Pelton, Francis, Kaplan), runner diameter, spiral casing size and generator diameter.
2. Length:
   • Unit spacing (centre to centre) = larger of spiral casing width or generator barrel diameter + clearance (about 1–2 m each side). For Pelton units, the distributor/manifold size governs.
   • Total length = number of units × unit spacing + erection (service) bay (about 1–1.5 unit spacing) + control/auxiliary room.
3. Width: governed by spiral casing and draft tube (Francis/Kaplan) or by turbine housing; plus space for main inlet valve, walkways, cable ducts and the overhead crane span.
4. Height:
   • Setting level of turbine from cavitation (Thoma σ) and tailwater level.
   • Draft tube depth below the runner (substructure).
   • Generator floor, and crane clearance high enough to lift the largest part (generator rotor) over other units.
5. Crane capacity and span (heaviest piece to lift).
6. Tailwater level and flood level: powerhouse floor above maximum flood; draft tube outlet submerged.
7. Auxiliaries: governors, control room, battery room, cooling water, transformers and switchyard space.
8. Site conditions: foundation geology, access road, space available (surface or underground).

Typical plan (Francis units)

 <-- erection --><-- unit --><-- unit --><ctrl>
     bay            bay 1       bay 2
+---------------+-----------+-----------+-----+
|               |  ( G1 )   |  ( G2 )   |     |
|  unloading,   |  spiral   |  spiral   | con-|
|  assembly     |  casing   |  casing   | trol|
|               |  MIV      |  MIV      |     |
+---------------+-----------+-----------+-----+
                  | draft tubes ->  tailrace

A typical run-of-river hydropower scheme diverts water from a river, carries it along a gentle slope to gain head, drops it through a penstock to the turbines, and returns it to the river.

 river
  |~~~|  diversion weir
  |===|--> intake --> gravel trap
  |~~~|               |
  |   |               v
  |   |          settling basin (desander)
  |   |               |
  |   |               v
  |   |     headrace canal / tunnel
  |   |               |
  |   |               v
  |   |     forebay / surge tank
  |   |               |
  |   |            penstock
  |   |               |
  |   |               v
  |   |<-- tailrace <-- POWERHOUSE --> switchyard
  |~~~|                     (turbine + generator)

Components

1. Diversion weir/dam: raises water level and diverts flow.
2. Intake: draws design flow, with trash rack and gate.
3. Gravel trap and settling basin: remove gravel and fine sand.
4. Headrace canal or tunnel: carries water at gentle slope.
5. Forebay (canal) or surge tank (tunnel): handles load changes.
6. Penstock: pressure pipe to the turbine.
7. Powerhouse: turbines, generators, controls.
8. Tailrace: returns water to the river.
9. Switchyard and transmission line: deliver power to the grid.

An intake is the structure at the head of a water conductor system that diverts the required design discharge from a river or reservoir into the canal, tunnel or penstock leading to the powerhouse, while keeping out debris, ice and as much sediment as possible.

Functions of an intake

1. Diversion of design flow: draws the required discharge (design discharge plus flushing flow) into the water conductor at all river stages, including the lean (dry) season.
2. Control of flow: gates at the intake regulate the flow and can shut it off completely for maintenance of the waterway or during floods.
3. Exclusion of floating debris: trash racks stop logs, leaves, boulders and ice from entering and damaging the turbines.
4. Exclusion of bed load: a raised sill, undersluice and gravel trap keep coarse sediment (gravel, boulders) out of the waterway.
5. Minimising head loss: a smooth bell-mouth entrance with low approach velocity (about 0.6–1.0 m/s through the racks) reduces entry losses.
6. Prevention of vortex and air entry: adequate submergence prevents vortices that would draw air into a pressure conduit.
7. Flood protection: limits the flow entering the waterway during floods so that the canal or tunnel is not overloaded; excess is spilled back to the river.

 River --> [Trash rack] -> [Gate] -> [Gravel trap]
                                        |
                                        v
                         Headrace canal / tunnel

In a run-of-river (RoR) plant there is little or no storage, so the intake must divert the design flow reliably in all seasons while keeping sediment out. The site is chosen by considering the following factors.

Hydraulic and river factors

1. Straight, stable reach: the river should be straight and stable for some distance upstream; if a bend is used, the intake is placed on the outer (concave) bank, where secondary currents carry bed load towards the inner bank and clear water flows towards the intake.
2. Assured water level: the site must allow the required water level in the dry season, either naturally or with a low weir, so that design flow can always be drawn.
3. Minimum sediment entry: avoid reaches with heavy bed load, braided channels or tributary confluences just upstream that bring large sediment.
4. Narrow river width: a narrow section with firm banks reduces the length and cost of the weir/diversion structure.
5. Flood levels: the intake and its gates must be safe during the design flood; high flood marks should be studied.

Geological and topographic factors

6. Sound foundation: bedrock or a firm foundation at shallow depth for the weir and intake structures.
7. Stable slopes: no landslides, debris flows or rock falls on the slopes around the intake.
8. Space for structures: enough flat area for the gravel trap, settling basin and approach to the headrace.

Layout and project factors

9. Head and alignment: the site should give the maximum gross head with the shortest and most economical headrace alignment to the forebay/surge tank.
10. Access and construction: reasonable road access, space for construction, river diversion during construction.
11. Environmental and social factors: downstream release (environmental flow), irrigation rights, fish migration, minimum land acquisition and resettlement.
12. Ice, debris and glacial hazards: in Nepal, sites exposed to GLOF and debris flows should be avoided or protected.

        River bend (plan)
     ~~~~~~~~~~~~~~~~~~~\
    flow ->              \   outer bank
     ~~~~~~~~~~~~~~~~~~\  \  [INTAKE] -> headrace
                        \  \
      inner bank (bed    \  \
      load deposits)      \  \

A settling basin (desander) is an enlarged chamber in the water conductor system where the flow velocity is reduced so that suspended sediment particles larger than a chosen size settle out before the water reaches the turbines. In Nepalese Himalayan rivers the silt load is very high and quartz-rich, so desanders are essential to prevent turbine abrasion.

Design principle

The basin works on the settling theory: a particle entering at the water surface at depth $h$ moves forward with the flow velocity $v$ and falls with its fall (settling) velocity $w$. It is trapped if it reaches the bed before the end of the basin.

$$\begin{aligned} \text{Time to settle } t &= \frac{h}{w}, \quad \text{time to pass } = \frac{L}{v} \\ \text{Condition: } \frac{L}{v} &\ge \frac{h}{w} \;\Rightarrow\; L \ge \frac{h\,v}{w} \end{aligned}$$

Width $B = \dfrac{Q}{v\,h}$. In practice turbulence reduces the effective fall velocity, so the length is increased using Velikanov's or Sakhmin's methods, e.g. Velikanov:

$$L = \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2}$$

where $\lambda$ depends on the required trap efficiency.

Design criteria

1. Particle size to be removed: usually 0.2–0.3 mm for medium heads; 0.1–0.15 mm for high-head Pelton plants (Nepal practice: 0.2 mm, or 0.15 mm for heads above about 250 m).
2. Trap efficiency: generally 90% or more for the design particle size.
3. Horizontal velocity: low enough to avoid resuspension; by Camp, $v = a\sqrt{d}$ cm/s ($a = 44$ for $0.1 < d < 1$ mm), usually 0.2–0.4 m/s.
4. Uniform flow distribution: gradual inlet transition (expansion about 1:5 to 1:8) and baffles/guide walls.
5. Depth: usually 3–6 m plus extra depth for sediment storage.
6. Flushing: sediment must be removed periodically or continuously (hydraulic flushing, Bieri system, Serpent system); flushing discharge must be available.
7. Number of chambers: at least two chambers so that one can be flushed while the other runs.
8. L/B ratio: generally 8–10 for good flow; a large width is split into several chambers.

Components of a settling basin

 PLAN
        inlet        settling zone        outlet
 from  ________   ___________________   ________
 intake  \     | |                   | |     /  to
 ------>  > ---| |   (two chambers)  | |--- <   headrace
 ______  /_____| |___________________| |_____\
        transition                  weir/transition

 L-SECTION
  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ WL
   \  inlet  |   settling zone        |outlet
    \________|________________________|
     sediment storage  ->  flushing canal/gate

1. Inlet zone (transition): gradually expanding section with guide walls or baffles that slows the water and spreads it uniformly over the basin width.
2. Settling zone: the main chamber of required length, width and depth where particles settle.
3. Sediment storage zone: extra depth below the settling zone, with sloping or hopper-shaped bed, where deposits collect.
4. Flushing system: flushing gates, conduits or channels with sufficient slope to return deposits to the river.
5. Outlet zone: converging transition and overflow weir that collect clear surface water into the headrace without disturbing the settled sediment.
6. Spillway/escape: a side spillway to release excess water and protect the basin.

Given: $Q = 12$ m³/s, $h = 3.5$ m, particle size $d = 0.2$ mm, fall velocity $w = 6$ cm/s $= 0.06$ m/s, turbulence coefficient $\lambda = 1.50$.

Method: horizontal velocity by Camp's formula, width from continuity, length by settling theory corrected for turbulence by Velikanov's formula.

1. Horizontal (flow-through) velocity

Camp: $v = a\sqrt{d}$ cm/s with $a = 44$ for $0.1 < d < 1$ mm.

$$\begin{aligned} v &= 44\sqrt{0.2} = 19.68\ \text{cm/s} = 0.197\ \text{m/s} \end{aligned}$$

2. Width of basin

$$\begin{aligned} B &= \frac{Q}{v\,h} = \frac{12}{0.1968 \times 3.5} = 17.42\ \text{m} \end{aligned}$$

3. Length by ideal settling theory (no turbulence)

$$\begin{aligned} t &= \frac{h}{w} = \frac{3.5}{0.06} = 58.3\ \text{s} \\ L_0 &= v\,t = \frac{h\,v}{w} = \frac{3.5 \times 0.1968}{0.06} = 11.48\ \text{m} \end{aligned}$$

4. Length considering turbulence (Velikanov)

$$\begin{aligned} L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\ &= \frac{1.5^2 \times 0.1968^2 \times (\sqrt{3.5} - 0.2)^2}{7.51 \times 0.06^2} \\ &= \frac{2.25 \times 0.03872 \times 2.7916}{0.02704} = 9.00\ \text{m} \end{aligned}$$

5. Adopted dimensions

The length must not be less than the ideal settling length, so the larger value (11.48 m) governs; adopt $L = 12$ m.

Answer: $v = 0.197$ m/s, $B \approx 17.4$ m, turbulent (Velikanov) length $= 9.0$ m; adopted basin about 12 m long × 17.5 m wide × 3.5 m deep (split into two chambers for flushing).

Note: the given $w = 6$ cm/s is high for 0.2 mm sand (real value about 2–2.5 cm/s); with a smaller $w$ the required length becomes much larger, so in practice the basin is made longer, with $L/B$ about 8–10 per chamber.

Given: $Q = 10$ m³/s, $h = 4.0$ m, particle size $d = 0.15$ mm, fall velocity $w = 6$ cm/s $= 0.06$ m/s, $\lambda = 1.50$.

Method: Camp's formula for flow velocity, continuity for width, settling theory with Velikanov's turbulence correction for length.

1. Horizontal velocity (Camp)

$v = a\sqrt{d}$ cm/s, $a = 44$ for $0.1 < d < 1$ mm:

$$v = 44\sqrt{0.15} = 17.04\ \text{cm/s} = 0.1704\ \text{m/s}$$

2. Width

$$B = \frac{Q}{v\,h} = \frac{10}{0.1704 \times 4.0} = 14.67\ \text{m}$$

3. Ideal settling length

$$\begin{aligned} t &= \frac{h}{w} = \frac{4.0}{0.06} = 66.7\ \text{s} \\ L_0 &= \frac{h\,v}{w} = \frac{4.0 \times 0.1704}{0.06} = 11.36\ \text{m} \end{aligned}$$

4. Length considering turbulence (Velikanov)

$$\begin{aligned} L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\ &= \frac{1.5^2 \times 0.1704^2 \times (2.0 - 0.2)^2}{7.51 \times 0.06^2} \\ &= \frac{2.25 \times 0.02904 \times 3.24}{0.02704} = 7.83\ \text{m} \end{aligned}$$

5. Adopted dimensions

Since the length cannot be less than the ideal settling length, $L_0 = 11.36$ m governs; adopt $L = 12$ m.

Answer: $v = 0.170$ m/s, $B \approx 14.7$ m, Velikanov length $= 7.83$ m; adopt a basin about 12 m long × 15 m wide × 4 m deep, divided into two chambers so that one can be flushed while the other operates.

Note: the given $w = 6$ cm/s is much larger than the real fall velocity of 0.15 mm sand (about 1.5 cm/s), which is why the computed length is short; with the real value the basin would be several times longer.

Given: $Q = 10$ m³/s, particle size $d = 0.25$ mm, trap efficiency $\eta = 95\%$.

Assumptions (not given): basin depth $h = 4.0$ m; water at about 20 °C; fall velocity of 0.25 mm quartz sand from the standard fall-velocity table, $w = 2.70$ cm/s $= 0.027$ m/s; Velikanov coefficient for 95% efficiency $\lambda = 1.65$ (Velikanov table: 90% → 1.28, 95% → 1.65, 98% → 2.05).

1. Horizontal velocity (Camp)

$$v = a\sqrt{d} = 44\sqrt{0.25} = 22\ \text{cm/s} = 0.22\ \text{m/s}$$

2. Width

$$B = \frac{Q}{v\,h} = \frac{10}{0.22 \times 4.0} = 11.36\ \text{m}$$

3. Ideal settling length (no turbulence)

$$\begin{aligned} t &= \frac{h}{w} = \frac{4.0}{0.027} = 148.1\ \text{s} \\ L_0 &= \frac{h\,v}{w} = \frac{4.0 \times 0.22}{0.027} = 32.6\ \text{m} \end{aligned}$$

4. Length considering turbulence (Velikanov)

Turbulent eddies keep particles in suspension longer, so the length is increased:

$$\begin{aligned} L &= \frac{\lambda^2 v^2 (\sqrt{h} - 0.2)^2}{7.51\,w^2} \\ &= \frac{1.65^2 \times 0.22^2 \times (2.0 - 0.2)^2}{7.51 \times 0.027^2} \\ &= \frac{2.7225 \times 0.0484 \times 3.24}{0.005475} = 77.98\ \text{m} \end{aligned}$$

5. Adopted dimensions

Answer: a two-chamber settling basin about 78 m long, 12 m wide (2 × 6 m) and 4 m deep (plus storage depth, inlet and outlet transitions). The turbulence effect more than doubles the length compared with the ideal settling length of 32.6 m. Each chamber has $L/B = 78/6 = 13$, which gives uniform flow.

A tunnel is an underground water conductor used as headrace or tailrace in hydropower projects, especially in the steep Himalayan terrain of Nepal.

Advantages of tunnelling

1. Shorter route: a tunnel can cut straight through a ridge or across a river loop, giving a much shorter waterway than a surface canal following contours (e.g. Kali Gandaki A, Middle Marsyangdi).
2. Safe from surface hazards: protected from landslides, rock falls, floods, debris flows and erosion, which often damage surface canals in Nepal.
3. Can carry pressure flow: a pressure tunnel allows drawing water from a fluctuating reservoir level and reduces penstock length.
4. Low maintenance: no damage from weather, cattle or people; little seepage in good rock.
5. Environment and land: minimal land acquisition and surface disturbance.
6. Good rock carries load: in sound rock the surrounding rock shares internal water pressure, so lining can be light or omitted.

Typical cross-section of a pressure tunnel

        surrounding rock mass
     .  .  .  .  .  .  .  .  .
   .     __________________    .
  .    /  rock bolt  ^       \   .
  .   / /  shotcrete ^  \  \  \  .
 .   | |   concrete lining | | |  .
 .   | |                    | | |  .
 .   | |   water (under    | | |  .
 .   | |    pressure)      | | |  .
  .   \ \                  / /  .
   .   \_\_______________/_/   .
    .  grout   invert  drain   .
      .  .  .  .  .  .  .  .

Key components:

• Excavated profile (circular or horseshoe/D-shape) in rock.
• Rock support: rock bolts, wire mesh and shotcrete to stabilise the opening.
• Lining: plain or reinforced concrete (steel lining near the powerhouse where cover is low) to reduce friction and leakage.
• Contact and consolidation grouting between lining and rock to fill voids and strengthen the rock.
• Invert (floor) for construction traffic, and drainage holes to relieve external water pressure.

A hydraulic tunnel is an underground conduit excavated through rock or soil to convey water, used in hydropower as headrace, tailrace, diversion or spillway tunnels. It may run free-flow (partly full, like an open channel) or pressure flow (full, under internal pressure).

Design features

1. Alignment: as straight and short as possible; through sound rock, avoiding faults, shear zones and weak strata; crossing weak zones at right angles; adequate rock cover (vertical and lateral) so that the rock weight resists the internal water pressure (Norwegian criterion: cover $C_{RM} \ge \dfrac{h_s \gamma_w F}{\gamma_r \cos\beta}$).
2. Type of flow: pressure tunnels need sound rock and lining for leakage control; free-flow tunnels need about 15–25% freeboard area above the water surface.
3. Shape of section:
   • Circular – best for high internal pressure, minimum perimeter, used for TBM tunnels.
   • Horseshoe / D-shaped (inverted-D) – easy drill-and-blast excavation and flat invert for traffic, common in Nepal.
   • Modified horseshoe – for external rock pressure.
4. Size (diameter): fixed by hydraulic and economic considerations—the economic diameter minimises the sum of construction cost and value of energy lost in friction; minimum size about 2.0–2.5 m for construction access. Usual velocities: 1–2.5 m/s in unlined, 2–4 m/s in lined tunnels.
5. Hydraulic losses: head loss by Manning, $h_f = \dfrac{n^2 V^2 L}{R^{4/3}}$; smooth lining reduces losses ($n \approx 0.012$–0.014 concrete, 0.03–0.04 unlined).
6. Lining and support: unlined in good rock; shotcrete and rock bolts in fair rock; concrete or steel lining in weak rock, low-cover zones and near the powerhouse. Lining reduces friction and leakage and resists water pressure.
7. Grouting and drainage: contact and consolidation grouting; drainage to relieve external water pressure when the tunnel is empty.
8. Slope: gentle slope (about 1:300–1:1000) towards the outlet for drainage and flushing; pressure tunnels kept below the minimum hydraulic grade line everywhere.
9. Transient (water hammer/surge) design: surge tank at the downstream end to limit pressure rise and allow quick load changes.
10. Rock trap and access: a rock trap near the downstream end and adits for construction, inspection and maintenance.

   Circular        Horseshoe        D-shaped
    .--.            .--.             .--.
   /    \          /    \           /    \
  |      |        |      |          |    |
   \    /          \    /           |    |
    '--'            \__/            |____|

The economic diameter of a tunnel (or penstock) is the diameter for which the total annual cost—the annual cost of construction (capital recovery + maintenance) plus the annual value of energy lost due to friction—is a minimum.

• A larger diameter gives lower velocity and smaller head loss, so less energy is lost, but construction cost (excavation, lining) is higher.
• A smaller diameter is cheaper to build but causes larger friction loss and loss of revenue every year.
• The economic diameter is the best balance between these two opposing costs.

Working it out

1. Annual construction cost. Cost of tunnel of length $L$ is roughly proportional to $D^2$ (excavation and lining). The annual cost is

$$\begin{aligned} C_1 = k_1 D^2 L \times (\text{capital recovery factor} + \text{O\&M rate}) \end{aligned}$$

2. Annual cost of energy lost. Head loss by Manning (for a full circular section $R = D/4$):

$$\begin{aligned} h_f &= \frac{n^2 V^2 L}{R^{4/3}}, \quad V = \frac{4Q}{\pi D^2} \\ \Rightarrow h_f &= 10.29\,\frac{n^2 Q^2 L}{D^{16/3}} \end{aligned}$$

Power lost $P_L = 9.81\,Q\,h_f\,\eta$ kW; annual energy lost $= P_L \times 8760 \times$ plant factor; its value at the energy rate $r$ gives

$$C_2 = \frac{k_2}{D^{16/3}}$$

3. Minimise total cost.

$$\begin{aligned} C &= C_1 + C_2 = k_1' D^2 + \frac{k_2}{D^{16/3}} \\ \frac{dC}{dD} &= 2k_1' D - \frac{16}{3}\frac{k_2}{D^{19/3}} = 0 \\ D_{eco} &= \left(\frac{8 k_2}{3 k_1'}\right)^{3/22} \end{aligned}$$

4. Graphical method (used in practice). Assume several trial diameters, compute for each the annual construction cost and annual loss cost, and plot them against $D$. The diameter at the lowest point of the total cost curve is the economic diameter.

 Annual cost
   ^   \                       / construction
   |    \                    /    cost
   |     \     total cost  /
   |      \__       ___ /
   |        \ ''-.-'' /
   |         \  min /'.
   |          '-..-'   '-. energy loss cost
   |              |       '--.__
   +--------------+---------------> D
                D_eco

5. Checks: velocity within the usual range (about 2–3 m/s in concrete-lined tunnels), minimum size for construction (about 2–2.5 m), and practical round size.

Function of forebay

A forebay is a small storage basin at the end of a headrace canal (free-flow), from which the penstock starts. Its functions are:

1. To provide a transition from open-channel flow to pressure flow in the penstock.
2. To act as a small balancing reservoir: supplies extra water when load suddenly increases and stores water (or spills it) when load decreases.
3. To absorb water-hammer/surge at the penstock entrance (free water surface) for the canal side.
4. To provide submergence above the penstock inlet to prevent vortex and air entry.
5. To act as a final settling basin and house trash racks, gates and a spillway for excess water.

Types of surge tank

A surge tank is an open-topped tank/shaft at the junction of a pressure tunnel and penstock that absorbs pressure rise and supplies water during load changes.

1. Simple surge tank: a vertical shaft of uniform section connected directly to the tunnel with an unrestricted opening. Simple and effective, but large size and slow damping of oscillations.
2. Restricted orifice (throttled) surge tank: a simple tank with a restricted orifice at its entry. The orifice loss quickly retards the tunnel flow, so the tank can be smaller and oscillations damp faster, but more water hammer passes into the tunnel.
3. Differential surge tank (Johnson type): a narrow central riser with ports opening into an outer large tank. The riser responds quickly while the outer tank provides storage; combines advantages of simple and restricted types.
4. Gallery (chamber) type surge tank: a narrow shaft with expansion chambers (galleries) at top and bottom—upper chamber limits up-surge, lower one prevents emptying in down-surge. Economical in rock (common in Nepal).
5. Inclined surge tank: shaft laid along a slope; acts like a larger area tank for a given height; suited to hillside.
6. Air-cushion (closed) surge chamber: an underground chamber with compressed air; water level fluctuations compress the air (used where topography does not allow an open shaft).

 Simple     Restricted   Differential   Gallery
  |  |        |  |       |  |  |  |     ==|  |==
  |  |        |  |       |  |  |  |       |  |
  |  |        |  |       |  |__|  |       |  |
 _|  |_      _|_ |_     _|___  ___|_    ==|  |==
 tunnel     orifice     riser+ports    chambers

A simple surge shaft is a vertical circular shaft at the end of a pressure tunnel. Its area is fixed for stable oscillations and its height must contain the maximum upsurge (on load rejection) and maximum downsurge (on load acceptance) without overflowing or letting air into the penstock.

            ___      Max upsurge level
           |   |  -- --------------------- +Z_max
  Reservoir|   |
 ~~~~~~~~~~|~~~|~~ static level (SWL)
   |       |   |  -- steady level (SWL - h_f)
   |       |   |  -- Min downsurge level  -Z_max
   |       | A |
   |_______|   |______________
   tunnel (A_t, L, h_f)      \ penstock -> turbine

Notation

$A_t$ = tunnel area, $L$ = tunnel length, $V_0$ = steady tunnel velocity, $Q_0 = A_t V_0$, $h_f$ = tunnel friction loss, $A_s$ = surge-shaft area, $g$ = 9.81 m/s².

1. Area of the shaft (Thoma criterion)

For the oscillations to be stable (damped) after a small load change, the area must be larger than the Thoma area:

$$A_{th} = \frac{L\,A_t}{2 g\, c\, H_0}, \qquad c = \frac{h_f}{V_0^2}$$

i.e.

$$A_{th} = \frac{L\,A_t\,V_0^2}{2 g\, h_f\, H_0}$$

where $H_0$ = net head on the turbine. A factor of safety of 1.5–1.8 is applied: $A_s = (1.5 \text{ to } 1.8)A_{th}$; diameter $D_s = \sqrt{4A_s/\pi}$.

2. Maximum surge (upsurge) – full load rejection

Neglecting friction, the dynamic equation of tunnel water and continuity of the shaft give simple harmonic oscillation:

$$\begin{aligned} \frac{L}{g}\frac{dV}{dt} &= -z, \qquad A_s\frac{dz}{dt} = A_t V \\ \Rightarrow\ \frac{d^2 z}{dt^2} &+ \frac{g A_t}{L A_s} z = 0 \end{aligned}$$

Solving with $V = V_0$ at $z = 0$:

$$Z_{max} = V_0\sqrt{\frac{L\,A_t}{g\,A_s}}, \qquad T = 2\pi\sqrt{\frac{L\,A_s}{g\,A_t}}$$

With friction, the upsurge is reduced, approximately (Jaeger/Johnson):

$$Z_{up} \approx Z_{max}\left(1 - \tfrac{2}{3}k + \tfrac{1}{9}k^2\right), \quad k = \frac{h_f}{Z_{max}}$$

3. Maximum downsurge – sudden load acceptance

For a sudden full-load demand, the downsurge below the static level is about

$$Z_{down} \approx h_f + (0.5 \text{ to } 1.0) Z_{max}$$

(often taken equal to $Z_{max}$ below the minimum steady level for a conservative design).

4. Height of the shaft

$$\begin{aligned} \text{Top level} &= \text{Max reservoir level} + Z_{up} + \text{freeboard (1–2 m)} \\ \text{Bottom level} &= \text{Min reservoir level} - Z_{down} - \text{submergence (2–3 m)} \end{aligned}$$

Height of shaft = top level − bottom level. The bottom must stay below the minimum downsurge so that air never enters the tunnel or penstock.

Objectives of a surge tank

A surge tank is an open shaft/chamber at the junction of the pressure tunnel and penstock. It is provided to:

1. Reduce water hammer pressure in the tunnel—the free surface reflects pressure waves so only the short penstock carries full water hammer.
2. Supply water quickly when turbine load increases (acts as a reservoir until tunnel flow accelerates).
3. Store water (rising level) when load is rejected, preventing excessive pressure rise.
4. Improve speed regulation of the turbine by reducing the effective length of the water column.

Factors governing the economical diameter of penstock

The economic diameter is the one giving minimum total annual cost (annual cost of the pipe + annual value of energy lost in friction). It depends on:

1. Discharge $Q$: larger flow needs larger diameter.
2. Head $H$ and pressure: higher heads need thicker plates, raising cost per unit diameter, so economic velocities are higher at high heads (smaller diameter relative to flow).
3. Length of penstock: head loss is proportional to length.
4. Cost of material and fabrication: steel price, plate thickness, welding, transport, anchor blocks and supports; higher cost favours smaller diameter.
5. Value (tariff) of energy: higher energy price favours a larger diameter to reduce losses.
6. Plant load factor and hours of operation: more running hours mean more energy lost through friction, favouring larger diameter.
7. Interest rate and life of penstock: affect the annual capital recovery cost.
8. Friction coefficient: pipe roughness (material, age, lining).
9. Water hammer and allowable velocity: larger velocity gives larger water hammer pressure and thicker pipe; erosion if silt-laden.
10. Transport and handling limits: maximum section size that can be carried on hill roads in Nepal.

Empirical formulas (e.g. USBR, Sarkaria, Fahlbusch) give a first estimate, refined by the cost-curve (graphical) method.

Thickness of penstock

A penstock is a thin cylinder under internal pressure. Consider a unit length of pipe of internal diameter $D$ cut along a diameter. The bursting force due to pressure $p$ is resisted by hoop tension in the two walls:

$$\begin{aligned} p \times D \times 1 &= 2\,\sigma\, t \times 1 \times \eta_j \\ \Rightarrow\ t &= \frac{p\,D}{2\,\sigma\,\eta_j} + t_c \end{aligned}$$

where $p = \gamma_w (H + h_{wh})$ is the design pressure (static head plus water hammer rise, typically 15–30%), $\sigma$ = allowable (design) stress of steel, $\eta_j$ = joint efficiency (0.85–0.95), $t_c$ = corrosion allowance (1.5–3 mm). A minimum thickness for handling is also used, e.g. $t_{min} = \dfrac{D + 500}{400}$ mm ($D$ in mm), not less than 6 mm.

Economic diameter of penstock

The economic diameter gives minimum total annual cost = annual cost of the penstock + annual value of energy lost by friction.

1. Pipe weight $\propto \pi D t$, and $t \propto pD$, so the annual cost of the penstock is

$$C_1 = k_1 D^2$$

2. Head loss (Darcy):

$$h_f = \frac{f L V^2}{2 g D} = \frac{8 f L Q^2}{\pi^2 g D^5}$$

Power lost $= 9.81\,Q\,h_f\,\eta$; annual value of lost energy

$$C_2 = \frac{k_2}{D^5}$$

3. Total annual cost and minimum:

$$\begin{aligned} C &= k_1 D^2 + \frac{k_2}{D^5} \\ \frac{dC}{dD} &= 2k_1 D - \frac{5k_2}{D^6} = 0 \\ D_{eco} &= \left(\frac{5k_2}{2k_1}\right)^{1/7} \end{aligned}$$

At the optimum, $2C_1 = 5C_2$, i.e. annual pipe cost is 2.5 times the annual cost of energy lost.

4. Graphical method: compute $C_1$ and $C_2$ for several trial diameters, plot them and their sum; the minimum of the total curve gives $D_{eco}$.

5. Empirical formulas for preliminary design ($P$ in kW, $H$ in m, $D$ in m):

• USBR (Bier): $D = 0.176\,(P/H)^{0.466}$
• Sarkaria: $D = 0.71\,P^{0.43}/H^{0.65}$
• Fahlbusch: $D = 0.52\,H^{-0.17}(P/H)^{0.43}$

Given: $Q = 6$ m³/s, $H = 60$ m, overall efficiency $\eta = 85\%$ (the "n" in the question is read as efficiency).

Power

$$\begin{aligned} P &= 9.81\,Q\,H\,\eta = 9.81 \times 6 \times 60 \times 0.85 \\ &= 3001.9\ \text{kW} \end{aligned}$$

i) USBR formula (Bier)

$D = 0.176\,(P/H)^{0.466}$, with $P$ in kW and $H$ in m:

$$\begin{aligned} \frac{P}{H} &= \frac{3001.9}{60} = 50.03 \\ D &= 0.176 \times 50.03^{0.466} = 0.176 \times 6.19 = 1.09\ \text{m} \end{aligned}$$

Velocity $V = \dfrac{6}{\pi(1.09)^2/4} = 6.43$ m/s.

ii) Sarkaria formula

$D = 0.71\,P^{0.43}/H^{0.65}$ ($P$ in kW, $H$ in m):

$$\begin{aligned} D &= \frac{0.71 \times 3001.9^{0.43}}{60^{0.65}} = \frac{0.71 \times 31.28}{14.31} \\ &= 1.55\ \text{m} \end{aligned}$$

Velocity $V = \dfrac{6}{\pi(1.55)^2/4} = 3.17$ m/s.

iii) Formula based on graphical solution

The empirical fit to the graphical (cost-curve) solution: $D = 1.12\,Q^{0.45}/H^{0.12}$ ($Q$ in m³/s, $H$ in m):

$$\begin{aligned} D &= \frac{1.12 \times 6^{0.45}}{60^{0.12}} = \frac{1.12 \times 2.240}{1.635} \\ &= 1.53\ \text{m} \end{aligned}$$

Velocity $V = \dfrac{6}{\pi(1.535)^2/4} = 3.24$ m/s.

Summary

Answer: $D_{USBR} = 1.09$ m, $D_{Sarkaria} = 1.55$ m, $D_{graphical} = 1.53$ m. The Sarkaria and graphical values agree closely and give a reasonable penstock velocity of about 3 m/s, so a diameter of about 1.5 m is adopted. (The USBR velocity form $V = 0.125\sqrt{2gH} = 4.29$ m/s gives $D = 1.33$ m.)

Given: $D = 1.5$ m $= 1500$ mm, static head $H = 250$ m, water hammer allowance 30%, design stress $\sigma = 120$ N/mm², joint efficiency $\eta_j = 0.85$.

Formula (thin cylinder, hoop stress):

$$t = \frac{p\,D}{2\,\sigma\,\eta_j}$$

Design pressure

$$\begin{aligned} H_d &= 1.30 \times 250 = 325\ \text{m} \\ p &= \gamma_w H_d = 9810 \times 325 = 3.188 \times 10^6\ \text{N/m}^2 \\ &= 3.188\ \text{N/mm}^2 \end{aligned}$$

Thickness

$$\begin{aligned} t &= \frac{3.188 \times 1500}{2 \times 120 \times 0.85} \\ &= \frac{4782.4}{204} = 23.44\ \text{mm} \end{aligned}$$

Answer: required plate thickness $t = 23.44$ mm, say 24 mm (adding a corrosion allowance of about 2 mm, a 26 mm plate would be used in practice).

Given: $D = 1.3$ m $= 130$ cm, pressure $p_0 = 20$ kg/cm², water hammer rise 20%, design stress $\sigma = 1050$ kg/cm², joint efficiency $\eta_j = 0.90$.

Formula (hoop stress in a thin pipe):

$$t = \frac{p\,D}{2\,\sigma\,\eta_j}$$

Design pressure

$$p = 1.20 \times 20 = 24\ \text{kg/cm}^2$$

Thickness

$$\begin{aligned} t &= \frac{24 \times 130}{2 \times 1050 \times 0.90} \\ &= \frac{3120}{1890} = 1.651\ \text{cm} = 16.51\ \text{mm} \end{aligned}$$

Answer: penstock thickness $t = 16.5$ mm, say 17 mm (plus about 2 mm corrosion allowance if required).

Given: $D = 1.20$ m $= 120$ cm, pressure $p_0 = 17.5$ kg/cm², water hammer rise 20%, design stress $\sigma = 1050$ kg/cm², joint efficiency $\eta_j = 0.87$.

Formula (hoop stress in a thin-walled pipe):

$$t = \frac{p\,D}{2\,\sigma\,\eta_j}$$

Design pressure

$$p = 1.20 \times 17.5 = 21\ \text{kg/cm}^2$$

Thickness

$$\begin{aligned} t &= \frac{21 \times 120}{2 \times 1050 \times 0.87} \\ &= \frac{2520}{1827} = 1.379\ \text{cm} = 13.79\ \text{mm} \end{aligned}$$

Answer: penstock thickness $t = 13.8$ mm, say 14 mm (a corrosion allowance of about 2 mm may be added, giving 16 mm).

A powerhouse with Francis turbines has three main levels: the substructure (draft tube and spiral casing in mass concrete), the intermediate structure (turbine and generator floor) and the superstructure (machine hall with overhead crane).

Typical layout (cross-section)

     ______________________________________
    |  roof truss / EOT crane rail         |
    |      [=== crane ===]                 |
    |                                      |
    |          _______                     |
    |         | GEN   |   generator floor  |
    |_________|_______|____________________|
    |          |shaft|    turbine floor    |
 penstock==>( spiral casing )  MIV         |
    |          \runner/                    |
    |           |    |  draft tube         |
    |__________/      \____________________|__ TWL
                \______\________~~~~~~~~~~~ tailrace

Plan: units in a row in the machine hall, with an erection (service) bay at one end at road level; control room, switchgear and transformer yard alongside.

Fixing the dimensions of the powerhouse

1. Length $= n \times$ unit spacing $+$ erection bay length $+$ end clearances.
   • Unit spacing is governed by the largest of: spiral casing width, draft tube outlet width, or generator barrel diameter, plus about 1.5–2 m for walkways and walls. Roughly 4–5 times the runner diameter $D$ for Francis units.
   • Erection bay ≈ 1 to 1.5 times the unit spacing, enough to assemble a rotor and runner.
2. Width is fixed by the spiral casing / generator size in the transverse direction (≈ 4–5 $D$) plus space for the main inlet valve (MIV) on the upstream side and governors, panels and walkways; draft tube length (≈ 4–5 $D$) fixes the substructure width.
3. Height:
   • Below the turbine centre line: draft tube depth (≈ 2.5–3 $D$) and the turbine setting $H_s$ below or above tailwater, fixed by the cavitation (Thoma) criterion.
   • Above the generator floor: height of the generator and exciter, plus the height needed for the crane to lift the rotor over other units, crane girder and roof (often 10–15 m above the generator floor).
4. Crane capacity is fixed by the heaviest part to be lifted (generator rotor or transformer).
5. Other: space for control room, auxiliaries, access, ventilation, and future units.

These empirical ratios (based on runner diameter $D$) are used at the feasibility stage; final dimensions come from manufacturer's drawings.