Water Conveyance Structures

Method used

Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

• Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
• Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
• Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

$$\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]$$

• The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
• Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Surge in the tunnel

$A_t = 50.27\ \text{m}^2$, $A_s = \frac{\pi}{4}(15)^2 = 176.71\ \text{m}^2$, $L = 4000$ m.

• $V = 25/50.27 = 0.497$ m/s
• $h_f = \dfrac{0.016 \times 4000 \times 0.497^2}{2(9.81)(8)} = 0.101$ m
• $Z_0 = 0.497\sqrt{\dfrac{4000 \times 50.27}{9.81 \times 176.71}} = 5.36$ m

(i) Maximum up-surge

$Z_{max} = 5.29$ m above reservoir level.

(ii) Maximum down-surge

First downswing $= 5.16$ m below reservoir level.

(iii) Water hammer pressure

Flow per penstock $= 25/2 = 12.5$ m³/s; $v = 12.5/3.142 = 3.979$ m/s. Closure time for full load rejection is taken as less than $2L/a = 2(500)/1600 = 0.625$ s (rapid closure):

$$h = \frac{a v}{g} = \frac{1600 \times 3.979}{9.81} = 649.0\ \text{m}$$

$p = \rho g h = 6.37\ \text{MPa}$ ($\approx 64.9\ \text{kg/cm}^2$).

(iv) Time of oscillation

$$T = 2\pi\sqrt{\frac{4000 \times 176.71}{9.81 \times 50.27}} = 237.9\ \text{s} \approx 4.0\ \text{min}$$

Answer: up-surge 5.29 m; down-surge 5.16 m; water hammer head 649.0 m (6.37 MPa); $T = 237.9$ s.

A forebay is a small reservoir at the end of the headrace (canal or tunnel) from which the penstock takes off. It stores water to meet sudden load demand, helps the transition from open-channel to pressure flow, and provides a place for the trash rack, spillway and flushing of sediment.

Design procedure

1. Decide the discharge and retention time: $Q$ = turbine discharge; retention time $t$ is usually 2-3 minutes at full load.
2. Volume of storage: $V=Q\,t$
3. Cross-section: limit the velocity in the forebay to $v\le0.2-0.3$ m/s, so that sediment settles and flow is calm: $A=Q/v$
4. Water depth from the submergence of the penstock intake:
   • velocity in penstock, $V_p=\dfrac{Q}{n\,\frac{\pi}{4}D^2}$
   • Gordon's formula $S=C\,V_p\sqrt D$ ($C=0.7245$ for symmetrical and 0.5434 for asymmetrical flow)
   • depth $H=S+D+$ bottom clearance (0.3-0.5 m)
5. Width and length: $B=\dfrac{A}{H}$, and $L=\dfrac{V}{A}$ (check $L/B$ is large enough, around 2-4, with a gradual expansion from the headrace).
6. Spillway: to pass the full design discharge when the turbines trip, $Q=C\,L_s\,h^{3/2}$ with $C\approx1.7$ for a broad crest and $h=0.3-0.5$ m.
7. Free board (0.5 m), bottom slope towards a flushing gate, trash rack at the intake and an air vent behind the penstock gate.
8. Check the entrance loss and the top of the forebay wall level = full supply level + surge + free board.

Sketch

PLAN
 headrace ====> \  forebay  /|==> penstock
        transition| rack     |
 ====>           /           |==> penstock
              spillway  |  flush gate
                        v to river

SECTION
 FSL ~~~~~~~~~~~~~~~~~~~~~~~~~~
      |      S      |rack|gate
      |             |####|  ___ penstock crown
 bed  |_____________|####|_/ D
        0.5 m clearance

The diameter of a power tunnel is chosen so that the net benefit (value of energy saved minus the cost of the tunnel) is highest. A small diameter is cheap but gives high velocity, high head loss and less energy; a large diameter costs more but loses less head.

Step by step procedure

1. Select several trial diameters $D_1, D_2, \dots$ (e.g. 3.0, 3.5, 4.0, 4.5, 5.0 m) for the design flow $Q$ and the tunnel length $L$.
2. Head loss for each: $h_f=\dfrac{f\,L}{D}\dfrac{V^2}{2g}$ or Manning, with $V=\dfrac{4Q}{\pi D^2}$, so $h_f\propto\dfrac{Q^2L}{D^5}$. Add the minor losses.
3. Power and energy loss: $P_{loss}=\rho g Q h_f\eta$ and the annual energy loss $E_{loss}=P_{loss}\times8760\times PF$ (PF = plant factor, and including the variation of Q through the flow-duration curve).
4. Annual cost of the lost energy: $C_E=E_{loss}\times\text{tariff}$ (loss of income). Add the cost of lost capacity.
5. Capital cost of the tunnel for each diameter: excavation, support, lining and grouting, usually of the form $C_c=k\,D^{m}$ (with $m$ about 1.2-2) per metre. Convert to an annual cost: $C_A=C_c\times CRF+\text{O\&M}$, where CRF is the capital recovery factor at the given interest rate and life.
6. Total annual cost: $C_T(D)=C_A(D)+C_E(D)$.
7. Optimum diameter: the diameter at which $C_T$ is minimum, i.e. where $\dfrac{dC_A}{dD}+\dfrac{dC_E}{dD}=0$; or the diameter at which the incremental benefit/cost ratio (extra energy value / extra annual cost) just exceeds 1. Plot $C_A$, $C_E$ and $C_T$ against $D$, and read the minimum.

If $C_c=kD^{m}$ and $C_E=a\,D^{-5}$, then

$$\frac{d}{dD}\left(k\,CRF\,D^{m}+aD^{-5}\right)=0\ \Rightarrow\ D_{opt}=\left(\frac{5a}{m\,k\,CRF}\right)^{\frac{1}{m+5}}$$

8. Check the practical limits: velocity (about 2.5-4 m/s for lined pressure tunnels, less for unlined), minimum construction diameter (about 2.5-3 m for TBM or drilling), geology and surge conditions.

Importance of lining in hydropower tunnels

1. Reduces friction loss: a smooth concrete surface has $n\approx0.013-0.015$ against $0.03-0.04$ for rough rock, so a smaller section can carry the same discharge with less head loss.
2. Structural support: carries the rock load and prevents falls of rock, slaking and collapse, especially in weak or jointed rock.
3. Resists internal water pressure in pressure tunnels: the lining (reinforced concrete or steel) transfers the pressure to the rock and prevents hydraulic fracturing.
4. Prevents leakage of water into the rock (loss of water, slope instability, damage to nearby structures) and prevents the entry of groundwater when the tunnel is empty.
5. Protects the rock from erosion and from deterioration, and prevents damage by high velocity flow.
6. Smooth flow reduces cavitation and vibration, and provides a durable surface for maintenance.

Types of lining

Grouting (contact and consolidation) is done behind the lining, and drainage holes relieve the external water pressure.

The shape of the tunnel cross-section is chosen according to the rock quality, the internal pressure, the construction method and the hydraulic efficiency.

 Circular      Horseshoe        D-shaped       Rectangular
   ___           ___            ______          _______
  /   \         /   \          /      \        |       |
 |     |       |     |        |        |       |       |
  \___/        |     |        |________|       |_______|
               \_____/

The economic diameter of a penstock is the diameter at which the total annual cost (cost of the pipe plus the cost of the energy lost in head loss) is the least. A large diameter costs more steel but loses less head.

Mathematical analysis

1. Cost of the penstock. The thickness is $t=\dfrac{p\,D}{2\sigma\eta}=\dfrac{\gamma H D}{2\sigma\eta}$ (neglecting the corrosion allowance), where $H$ is the head including water hammer, $\sigma$ is the allowable stress and $\eta$ is the joint efficiency. The weight of steel per unit length is $W=\pi D\,t\,\gamma_s\propto D^2$. With the unit cost $c_s$ per unit weight and the annual charge rate $r$ (interest + depreciation + maintenance):

$$C_1=r\,c_s\,\pi\gamma_s\frac{\gamma H}{2\sigma\eta}\,L\,D^2=K_1\,D^2$$

2. Cost of the energy lost. The head loss in a pipe of length $L$ carrying $Q$ is

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

The power lost is $P=\gamma Q h_f\eta_{pl}$ and the annual value of the lost energy (with plant factor PF and unit rate $c_e$):

$$C_2=\gamma Q\,\eta_{pl}\,c_e\,(8760\,PF)\frac{8fLQ^2}{\pi^2gD^5}=\frac{K_2\,Q^3}{D^5}$$

3. Total annual cost and optimisation.

$$C=K_1D^2+\frac{K_2Q^3}{D^5}$$ $$\frac{dC}{dD}=2K_1D-\frac{5K_2Q^3}{D^6}=0\ \Rightarrow\ D_{econ}=\left(\frac{5K_2}{2K_1}\right)^{1/7}Q^{3/7}$$

Thus $D_{econ}\propto Q^{3/7}$. At the optimum, the cost of energy lost is $\frac{2}{5}$ of the cost of the pipe: $C_2=\frac{2}{5}C_1$.

4. Practical use. Many empirical formulas relate $D$ to $Q$ and $H$, but all show the same trend $D\propto Q^{3/7}$ found above. After the analysis, the diameter is rounded to a standard plate/pipe size, and the velocity (about 3-6 m/s), the head loss and the water hammer are checked.

Method used

Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

• Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
• Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
• Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

$$\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]$$

• The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
• Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Surge in the tunnel

$A_t = \frac{\pi}{4}(8)^2 = 50.27\ \text{m}^2$, $A_s = \frac{\pi}{4}(20)^2 = 314.16\ \text{m}^2$, $L = 4000$ m.

• $V = 45/50.27 = 0.895$ m/s
• $h_f = \dfrac{0.016 \times 4000 \times 0.895^2}{2(9.81)(8)} = 0.327$ m
• $Z_0 = 0.895\sqrt{\dfrac{4000 \times 50.27}{9.81 \times 314.16}} = 7.23$ m

a) Maximum up-surge

$Z_{max} = 7.02$ m above reservoir level.

b) Maximum down-surge

First downswing $= 6.63$ m below reservoir level.

c) Water hammer pressure

Each penstock carries $45/3 = 15$ m³/s. $A_p = \frac{\pi}{4}(2)^2 = 3.142\ \text{m}^2$, so $v = 15/3.142 = 4.775$ m/s.

Full load rejection means the gates close suddenly (closure time $\le 2L/a = 2(500)/1400 = 0.71$ s), so the Joukowsky rise is

$$h = \frac{a v}{g} = \frac{1400 \times 4.775}{9.81} = 681.4\ \text{m}$$

$p = \rho g h = 6.69\ \text{MPa}$ ($\approx 68.1\ \text{kg/cm}^2$).

d) Time of oscillation

$$T = 2\pi\sqrt{\frac{4000 \times 314.16}{9.81 \times 50.27}} = 317.2\ \text{s} \approx 5.3\ \text{min}$$

(The pressure wave in the penstock itself has a period $4L_p/a = 4(500)/1400 = 1.43$ s.)

Answer: up-surge 7.02 m; down-surge 6.63 m; water hammer head 681.4 m (6.69 MPa); $T = 317.2$ s.

Method used

Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

• Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
• Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
• Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

$$\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]$$

• The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
• Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Surge in the tunnel

$A_t = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $A_s = \frac{\pi}{4}(30)^2 = 706.86\ \text{m}^2$, $L = 7000$ m.

• $V = 60/78.54 = 0.764$ m/s
• $h_f = \dfrac{0.016 \times 7000 \times 0.764^2}{2(9.81)(10)} = 0.333$ m
• $Z_0 = 0.764\sqrt{\dfrac{7000 \times 78.54}{9.81 \times 706.86}} = 6.80$ m

(i) Maximum up-surge and down-surge

• Up-surge $= 6.58$ m above reservoir level.
• Down-surge (first downswing) $= 6.19$ m below reservoir level.

(ii) Water hammer pressure

Discharge per penstock $= 60/3 = 20$ m³/s; $A_p = 3.142\ \text{m}^2$; $v = 20/3.142 = 6.366$ m/s.

$$h = \frac{a v}{g} = \frac{1800 \times 6.366}{9.81} = 1168.1\ \text{m}$$

$p = 11.46\ \text{MPa}$ ($\approx 116.8\ \text{kg/cm}^2$), for instantaneous (rapid) closure.

(iii) Time of oscillation

$$T = 2\pi\sqrt{\frac{7000 \times 706.86}{9.81 \times 78.54}} = 503.5\ \text{s} \approx 8.4\ \text{min}$$

Answer: up-surge 6.58 m; down-surge 6.19 m; water hammer head 1168.1 m (11.46 MPa); $T = 503.5$ s.

Method used

Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

• Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
• Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
• Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

$$\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]$$

• The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
• Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Calculation

$A_t = \frac{\pi}{4}(4)^2 = 12.566\ \text{m}^2$, $A_s = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $L = 4000$ m, $f = 0.018$.

• $V = 20/12.566 = 1.592$ m/s
• $h_f = \dfrac{0.018 \times 4000 \times 1.592^2}{2(9.81)(4)} = 2.32$ m
• $Z_0 = 1.592\sqrt{\dfrac{4000 \times 12.566}{9.81 \times 78.54}} = 12.86$ m

Instantaneous closure of the turbine gates means complete rejection of the discharge, so the water rises in the tank. The penstock (3.2 m, 700 m) lies downstream of the tank, so it does not affect the tank oscillation.

Maximum height of surge tank

From the upsurge equation with friction: $Z_{max} = 11.36$ m above reservoir level (the frictionless value is 12.86 m).

The tank must therefore extend at least 11.4 m above the reservoir water level (plus a freeboard of about 1 to 2 m). The first downswing is 9.36 m below reservoir level.

Time period

$$T = 2\pi\sqrt{\frac{4000 \times 78.54}{9.81 \times 12.566}} = 317.2\ \text{s} \approx 5.3\ \text{min}$$

Answer: maximum upsurge height $= 11.36$ m above reservoir level; $T = 317.2$ s.

Method used

Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$.

• Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$.
• Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$.
• Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity:

$$\frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right]$$

• The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level).
• Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$.

Calculation

$A_t = \frac{\pi}{4}(4.5)^2 = 15.904\ \text{m}^2$, $A_s = \frac{\pi}{4}(10)^2 = 78.54\ \text{m}^2$, $L = 2500$ m, $f = 0.02$.

• $V = 25/15.904 = 1.572$ m/s
• $h_f = \dfrac{0.02 \times 2500 \times 1.572^2}{2(9.81)(4.5)} = 1.40$ m
• $Z_0 = 1.572\sqrt{\dfrac{2500 \times 15.904}{9.81 \times 78.54}} = 11.29$ m

Instantaneous closure means full rejection, so the water rises in the tank. The penstock is downstream of the tank and does not change the tank oscillation.

Maximum height of surge tank

With friction, $Z_{max} = 10.38$ m above reservoir level (frictionless value 11.29 m). The tank should rise at least 10.4 m above the reservoir level, plus freeboard of about 1 to 2 m. The first downswing is 9.01 m below reservoir level.

Time period

$$T = 2\pi\sqrt{\frac{2500 \times 78.54}{9.81 \times 15.904}} = 222.9\ \text{s} \approx 3.7\ \text{min}$$

Answer: maximum upsurge height $= 10.38$ m above reservoir level; $T = 222.9$ s.

Design considerations of a forebay

• Storage: enough volume for the retention time (2-3 min) to meet a sudden load increase until the headrace flow adjusts.
• Low velocity (0.2-0.3 m/s) for settling of the remaining sediment and calm flow at the intake.
• Adequate submergence of the penstock intake below the minimum water level to avoid vortices and air entry (Gordon's rule), with a trash rack and a smooth bell-mouth.
• Spillway with a capacity of the design flow, to pass the water on load rejection, with a safe discharge channel.
• Flushing/desilting arrangement at the bottom (flushing gate and slope) and a bottom clearance of about 0.5 m.
• Stable foundation, water-tightness (lining), free board 0.5 m and access for maintenance; located on firm ground, close to the powerhouse, and protected from slides.

Design for the given data

Given: $Q=12.0$ m³/s; 2 penstocks of diameter 2.0 m; retention (detention) time 3 min; limiting velocity in the forebay 0.20 m/s.

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=12.0\times(3\times60)=2160\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{12.0}{0.20}=60.0\ \text{m}^2$$

Step 3: Depth of water (submergence of the penstock intake)

Velocity in each penstock: $V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=1.910$ m/s. Minimum submergence to avoid vortices (Gordon, symmetrical approach):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times1.910\times\sqrt{2.0}=1.96\ \text{m}$$

Water depth = $S+D+$ bottom clearance 0.5 m $=1.96+2.0+0.5=4.46$ m, adopt $H=4.5$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{60.0}{4.5}=13.33\ \text{m},\qquad L=\frac{V}{A}=\frac{2160}{60.0}=36.0\ \text{m}$$

Adopt $B=13.5$ m and $L=36$ m (the volume is then $B\,L\,H=2187$ m³ $\ge V$).

PLAN
 headrace        forebay                 penstocks
 tunnel/canal  +------------------+
 ====> \       |                  |==> penstock 1
        \ trans| trash rack       |
 ====> /       |                  |==> penstock 2
              +--+-spillway-+-flush+
                   |          |
                   v          v  to river (spillway/flushing)

SECTION
 canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~
                  |        S (submergence)
                  |   rack  ____ crown
                  |   ####  \   penstock
 flushing  ______ |___####____\_D__
 bottom clearance 0.5 m

Answer: forebay about 36 m long x 13.5 m wide x 4.5 m deep (water), with a volume of 2160 m³.

Given: $Q=14.0$ m³/s; 2 penstocks of diameter 2.0 m and length 500 m; retention (detention) time 3 min; limiting velocity in the forebay 0.20 m/s.

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=14.0\times(3\times60)=2520\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{14.0}{0.20}=70.0\ \text{m}^2$$

Step 3: Depth of water (submergence of the penstock intake)

Velocity in each penstock: $V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=2.228$ m/s. Minimum submergence to avoid vortices (Gordon, symmetrical approach):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times2.228\times\sqrt{2.0}=2.28\ \text{m}$$

Water depth = $S+D+$ bottom clearance 0.5 m $=2.28+2.0+0.5=4.78$ m, adopt $H=4.8$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{70.0}{4.8}=14.58\ \text{m},\qquad L=\frac{V}{A}=\frac{2520}{70.0}=36.0\ \text{m}$$

Adopt $B=15.0$ m and $L=36$ m (the volume is then $B\,L\,H=2592$ m³ $\ge V$).

The penstock length (500 m) is not needed for sizing the forebay by this method; it matters for the penstock head loss and the water hammer.

Step 5: Spillway of the forebay

If the turbines trip, the whole discharge must spill over the forebay spillway (side-channel type) while the water rises by $h=0.5$ m:

$$Q=C\,L_s\,h^{3/2}\ \Rightarrow\ L_s=\frac{Q}{C\,h^{3/2}}=\frac{14.0}{1.7\times0.5^{3/2}}=23.3\ \text{m}$$

($C=1.7$ for a broad-crested weir.) Adopt $L_s=24$ m, placed along the side wall (or folded), with a spill channel to the river.

PLAN
 headrace        forebay                 penstocks
 tunnel/canal  +------------------+
 ====> \       |                  |==> penstock 1
        \ trans| trash rack       |
 ====> /       |                  |==> penstock 2
              +--+-spillway-+-flush+
                   |          |
                   v          v  to river (spillway/flushing)

SECTION
 canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~
                  |        S (submergence)
                  |   rack  ____ crown
                  |   ####  \   penstock
 flushing  ______ |___####____\_D__
 bottom clearance 0.5 m

Answer: forebay about 36 m long x 15.0 m wide x 4.8 m deep (water), with a volume of 2520 m³.

Given: $Q=14.5$ m³/s; 2 penstocks of diameter 1.8 m; retention (detention) time 3 min; limiting velocity in the forebay 0.22 m/s.

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=14.5\times(3\times60)=2610\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{14.5}{0.22}=65.9\ \text{m}^2$$

Step 3: Depth of water (submergence of the penstock intake)

Velocity in each penstock: $V_p=\dfrac{Q}{2\times\frac{\pi}{4}D^2}=2.849$ m/s. Minimum submergence to avoid vortices (Gordon, symmetrical approach):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times2.849\times\sqrt{1.8}=2.77\ \text{m}$$

Water depth = $S+D+$ bottom clearance 0.5 m $=2.77+1.8+0.5=5.07$ m, adopt $H=5.1$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{65.9}{5.1}=12.92\ \text{m},\qquad L=\frac{V}{A}=\frac{2610}{65.9}=39.6\ \text{m}$$

Adopt $B=13.0$ m and $L=40$ m (the volume is then $B\,L\,H=2652$ m³ $\ge V$).

PLAN
 headrace        forebay                 penstocks
 tunnel/canal  +------------------+
 ====> \       |                  |==> penstock 1
        \ trans| trash rack       |
 ====> /       |                  |==> penstock 2
              +--+-spillway-+-flush+
                   |          |
                   v          v  to river (spillway/flushing)

SECTION
 canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~
                  |        S (submergence)
                  |   rack  ____ crown
                  |   ####  \   penstock
 flushing  ______ |___####____\_D__
 bottom clearance 0.5 m

Answer: forebay about 40 m long x 13.0 m wide x 5.1 m deep (water), with a volume of 2610 m³.

Assumptions: a single penstock (as only one diameter is given); limiting velocity 0.2 m/s.

Given: $Q=20.0$ m³/s; 1 penstock of diameter 2.2 m and length 300 m; retention (detention) time 3 min; limiting velocity in the forebay 0.20 m/s (assumed, as it is not given).

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=20.0\times(3\times60)=3600\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{20.0}{0.20}=100.0\ \text{m}^2$$

Step 3: Depth of water (submergence of the penstock intake)

Velocity in each penstock: $V_p=\dfrac{Q}{1\times\frac{\pi}{4}D^2}=5.261$ m/s. Minimum submergence to avoid vortices (Gordon, symmetrical approach):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times5.261\times\sqrt{2.2}=5.65\ \text{m}$$

Water depth = $S+D+$ bottom clearance 0.5 m $=5.65+2.2+0.5=8.35$ m, adopt $H=8.4$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{100.0}{8.4}=11.90\ \text{m},\qquad L=\frac{V}{A}=\frac{3600}{100.0}=36.0\ \text{m}$$

Adopt $B=12.0$ m and $L=36$ m (the volume is then $B\,L\,H=3629$ m³ $\ge V$).

The penstock length (300 m) is not needed for sizing the forebay by this method; it matters for the penstock head loss and the water hammer.

PLAN
 headrace        forebay                 penstocks
 tunnel/canal  +------------------+
 ====> \       |                  |==> penstock 1
        \ trans| trash rack       |
 ====> /       |                  |==> penstock 2
              +--+-spillway-+-flush+
                   |          |
                   v          v  to river (spillway/flushing)

SECTION
 canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~
                  |        S (submergence)
                  |   rack  ____ crown
                  |   ####  \   penstock
 flushing  ______ |___####____\_D__
 bottom clearance 0.5 m

Answer: forebay about 36 m long x 12.0 m wide x 8.4 m deep (water), with a volume of 3600 m³.

Assumptions: a single penstock; limiting velocity 0.2 m/s (usual limit).

Given: $Q=20.0$ m³/s; 1 penstock of diameter 2.2 m and length 300 m; retention (detention) time 3 min; limiting velocity in the forebay 0.20 m/s (assumed, as it is not given).

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=20.0\times(3\times60)=3600\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{20.0}{0.20}=100.0\ \text{m}^2$$

Step 3: Depth of water (submergence of the penstock intake)

Velocity in each penstock: $V_p=\dfrac{Q}{1\times\frac{\pi}{4}D^2}=5.261$ m/s. Minimum submergence to avoid vortices (Gordon, symmetrical approach):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times5.261\times\sqrt{2.2}=5.65\ \text{m}$$

Water depth = $S+D+$ bottom clearance 0.5 m $=5.65+2.2+0.5=8.35$ m, adopt $H=8.4$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{100.0}{8.4}=11.90\ \text{m},\qquad L=\frac{V}{A}=\frac{3600}{100.0}=36.0\ \text{m}$$

Adopt $B=12.0$ m and $L=36$ m (the volume is then $B\,L\,H=3629$ m³ $\ge V$).

The penstock length (300 m) is not needed for sizing the forebay by this method; it matters for the penstock head loss and the water hammer.

PLAN
 headrace        forebay                 penstocks
 tunnel/canal  +------------------+
 ====> \       |                  |==> penstock 1
        \ trans| trash rack       |
 ====> /       |                  |==> penstock 2
              +--+-spillway-+-flush+
                   |          |
                   v          v  to river (spillway/flushing)

SECTION
 canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~
                  |        S (submergence)
                  |   rack  ____ crown
                  |   ####  \   penstock
 flushing  ______ |___####____\_D__
 bottom clearance 0.5 m

Answer: forebay about 36 m long x 12.0 m wide x 8.4 m deep (water), with a volume of 3600 m³.

Assumptions: a single penstock; limiting velocity 0.2 m/s (usual limit).

Given: $Q=15.0$ m³/s; 1 penstock of diameter 2.0 m and length 500 m; retention (detention) time 4 min; limiting velocity in the forebay 0.20 m/s (assumed, as it is not given).

Step 1: Volume of the forebay

The forebay stores the water for the retention time:

$$V=Q\,t=15.0\times(4\times60)=3600\ \text{m}^3$$

Step 2: Cross-sectional area from the velocity limit

$$A=\frac{Q}{v}=\frac{15.0}{0.20}=75.0\ \text{m}^2$$

Step 3: Depth of water (submergence of the penstock intake)

Velocity in each penstock: $V_p=\dfrac{Q}{1\times\frac{\pi}{4}D^2}=4.775$ m/s. Minimum submergence to avoid vortices (Gordon, symmetrical approach):

$$S=0.7245\,V_p\sqrt{D}=0.7245\times4.775\times\sqrt{2.0}=4.89\ \text{m}$$

Water depth = $S+D+$ bottom clearance 0.5 m $=4.89+2.0+0.5=7.39$ m, adopt $H=7.4$ m.

Step 4: Width and length

$$B=\frac{A}{H}=\frac{75.0}{7.4}=10.14\ \text{m},\qquad L=\frac{V}{A}=\frac{3600}{75.0}=48.0\ \text{m}$$

Adopt $B=10.5$ m and $L=48$ m (the volume is then $B\,L\,H=3730$ m³ $\ge V$).

The penstock length (500 m) is not needed for sizing the forebay by this method; it matters for the penstock head loss and the water hammer.

PLAN
 headrace        forebay                 penstocks
 tunnel/canal  +------------------+
 ====> \       |                  |==> penstock 1
        \ trans| trash rack       |
 ====> /       |                  |==> penstock 2
              +--+-spillway-+-flush+
                   |          |
                   v          v  to river (spillway/flushing)

SECTION
 canal bed ____ WL ~~~~~~~~~~~~~~~~~~~~~~
                  |        S (submergence)
                  |   rack  ____ crown
                  |   ####  \   penstock
 flushing  ______ |___####____\_D__
 bottom clearance 0.5 m

Answer: forebay about 48 m long x 10.5 m wide x 7.4 m deep (water), with a volume of 3600 m³.

Tunnelling methods in hydropower projects

1. Drill and blast (conventional method): the most common method in Nepal. The cycle is drilling holes, charging with explosives, blasting, ventilation, mucking (removal of rock), scaling and installation of support. Large sections are excavated in stages: top heading then benching.
2. Tunnel boring machine (TBM): a rotary cutter-head that excavates the full circular section and installs the precast segments or supports; fast and gives a smooth surface, but costly and not suitable in squeezing or highly variable ground. Used for long, uniform tunnels.
3. NATM / sprayed-concrete method (New Austrian Tunnelling Method): the excavation is supported at once by shotcrete, rock bolts and mesh, and the rock itself carries the load; the deformation is monitored.
4. Mechanical excavation (road-header) in soft rock, and cut-and-cover for shallow sections.
5. Adits and shafts are used to open extra faces; raise boring is used for shafts.

Why tunnel supports are provided

• To prevent the fall of rock and collapse of the opening, and so protect workers and machines.
• To control the deformation and the stress redistribution, particularly in weak, jointed or squeezing rock and fault zones.
• To stop the loosening of the rock and the inflow of water.
• To provide a base for the final lining.

How the supports are realized

1. Scaling of loose rock after each blast.
2. Shotcrete (5-15 cm) with wire mesh or fibre is sprayed on the crown and walls straight after excavation.
3. Rock bolts (grouted or friction), 2-4 m long at 1-2 m spacing, are installed through the shotcrete to bind the rock.
4. Steel ribs/arches (I-sections or lattice girders) with lagging in poor rock, installed close to the face and encased in shotcrete.
5. Pre-support (spiles, forepoling, pipe umbrella) in very weak ground and fault zones, ahead of the face.
6. Final concrete lining (and steel lining for pressure sections) after the excavation is finished, followed by contact and consolidation grouting.

The type and amount of support follow the rock mass classification (Q-system, RMR) and are adapted to the rock seen at the face.

Tunnelling methods in hydropower projects

1. Drill and blast: drilling, charging, blasting, ventilation, mucking, scaling and support. Used most in Nepal. Large faces are excavated by top heading and benching.
2. Tunnel boring machine (TBM): full-face rotary excavation with segmental lining; high speed and smooth tunnel, but costly and needs uniform ground and a long drive .
3. NATM: excavation with immediate shotcrete, rock bolts and mesh, and monitoring of deformation, which lets the rock carry part of the load.
4. Road-header or hydraulic breakers in soft rock, and cut-and-cover at shallow portals.

Purpose of shotcreting

Shotcrete is concrete sprayed at high speed onto the excavated rock surface.

• Seals the surface, preventing weathering, slaking and loosening of the joints.
• Gives immediate support and takes over the rock load together with bolts and mesh.
• Fills the cracks and bonds the rock blocks into a ring (arch action).
• Smoothens the surface and reduces overbreak and hydraulic roughness (as a lining of non-pressure tunnels).
• Gives quick, flexible support with little cost.

Procedure of shotcreting

1. Scale the loose rock and wash the surface with air-water jet to remove dust and loose particles.
2. Install mesh or add steel/synthetic fibres and set the thickness gauge pins.
3. Prepare the mix (cement, aggregate up to 10 mm, admixtures, accelerator) by the dry-mix (water added at the nozzle) or wet-mix (pumped wet concrete) process.
4. Spray with the nozzle held 1-1.5 m from the surface and at right angles to it, moving in small circles. Spray from the walls upwards to the crown, in layers of 5-10 cm; apply the second layer after the first has set.
5. Control the rebound (material that falls away) and remove it; do not reuse it.
6. Cure the shotcrete by spraying water or a curing compound for several days.
7. Quality tests: thickness checks, strength cores/panel tests, and bond tests.

Types of tunnel supports

  Rock bolts          Steel rib          Shotcrete + mesh
  \  |  /          ___________            ___________
 --\ | /--        /  ribs +   \          / sprayed  \
  /-+-+-\        |  lagging    |        | concrete    |
 /  | |  \       |             |        |   layer     |

1. Rock bolts: steel bars (fully grouted or friction type) 2-5 m long, installed in a pattern. They bind the loose blocks into the sound rock and form a reinforced ring.
2. Shotcrete: sprayed concrete (with mesh or fibres) that seals and supports the surface.
3. Steel ribs/arches: I-section or lattice girders, with lagging (timber or steel plates) or shotcrete in between; used for weak rock and fault zones; closed rings used for squeezing ground.
4. Concrete lining (plain or reinforced; cast-in-place or precast): the permanent support.
5. Wire mesh and forepoling / spiling / pipe-roof (pre-support) ahead of the face in very weak ground.
6. Timber supports for small, temporary works (rarely in large tunnels).

Necessity of supports

They keep the opening stable and safe, prevent fall of rock, control deformation and water inflow, and carry the load until the permanent lining is placed. The type is selected by the rock quality (Q-system/RMR): sound rock needs little or no support, jointed rock needs bolts and shotcrete, and poor rock needs ribs and pre-support.

Hydraulic design of a tunnel

It is the selection of the shape, size and slope of the tunnel so that the design discharge flows with an acceptable velocity and head loss, and so that the cost is the minimum.

1. Determine the design discharge $Q$ and select the type: free-flow (non-pressure) or pressure tunnel.
2. Choose the velocity: about 1.5-2.5 m/s for unlined and 2.5-4 m/s for lined tunnels (up to 5-6 m/s for pressure tunnels).
3. Size: $A=Q/V$; for circular, $D=\sqrt{4A/\pi}$. The economic diameter is found by optimisation (cost of the tunnel against the value of the energy lost).
4. Head loss: Manning $V=\frac{1}{n}R^{2/3}S^{1/2}$ ($n=0.013-0.015$ concrete, $0.03-0.04$ unlined rock) or Darcy-Weisbach; add the local losses (entry, bends, transitions).
5. For a free-flow tunnel, provide an air space of 15-20% of the area, a bed slope of 1:500-1:1000, and keep $Fr$ below 0.7 (not near 1) to avoid unstable flow.
6. For a pressure tunnel, the hydraulic grade line (including water hammer/surge) must stay above the crown, with a minimum rock cover.

Selection of tunnel alignment

• Shortest and straightest route possible (lowest cost and head loss), with gentle bends.
• Good geology: avoid faults, shear zones, weak and water-bearing rock and landslide areas; cross the weak zones at right angles; avoid squeezing and swelling rock.
• Adequate rock cover so that the internal pressure cannot burst the rock: the vertical rock cover must satisfy the Norwegian criterion $C_{RM}\ge\dfrac{h_s\gamma_w F}{\gamma_r\cos\beta}$ ($h_s$ = static head, $F\approx1.3$, $\beta$ = slope of the ground).
• Portals and adits: stable slopes at the portals and adits at suitable intervals for faster construction; access roads.
• Ground water and inflow: avoid high water pressure and heavy inflow zones.
• Connection with the intake, surge tank and powerhouse so that the layout (surge shaft and penstock) is simple, with a sufficient head and low cost.
• Environment and safety: avoid damage to settlements and existing structures.

Hydraulic design considerations of the surge tank

• Location: as close to the powerhouse as possible, on stable ground at a level high enough that the maximum surge level stays below the tank top, and the minimum level stays above the tunnel crown.
• Type: simple, restricted orifice, differential (Johnson), or with an upper/lower chamber, chosen by the head, topography and the required reduction of the surge.
• Size: the cross-section must be larger than the Thoma area for stability under small load changes:

$$A_s\ge A_{th}=\frac{L\,A_t}{2g\,\alpha\,(H_0-h_f)}$$

• Surge levels: maximum upsurge on full load rejection must not overflow, and minimum downsurge on full-load demand must not expose the tunnel crown (air entry) or the tank bottom; the oscillation period $T=2\pi\sqrt{LA_s/(gA_t)}$ should be checked with the governor.
• Connection: the area of the connecting orifice and the shaft should be sized to give small head loss, but enough damping.
• Provide an overflow, air vents, and access.

Hydraulic design considerations of a pressurised tunnel

• Diameter and velocity: from the economic diameter; velocity usually 2.5-4 m/s.
• Friction and other losses: Manning/Darcy, with $n$ of lining; plus entry, bend and transition losses.
• Internal pressure: the maximum internal pressure includes the static head and the surge/water-hammer rise.
• Minimum rock cover: the cover should resist the internal pressure (about $C_{rm}\ge\dfrac{h_s\gamma_w F}{\gamma_r\cos\beta}$); otherwise the section is steel-lined.
• Hydraulic grade line should stay above the crown at all times, in the transient too, so as to avoid vacuum/negative pressure.
• Lining and grouting: concrete lining for smoothness, steel lining where the rock is weak or the cover is low; contact and consolidation grouting; drainage.
• Protection: air vents and an emergency gate at the intake, and a surge tank at the downstream end to reduce the water hammer in the tunnel.

Functions of a surge tank

1. Reduces water hammer: it reflects the pressure waves, so the long headrace tunnel is protected from high pressure rise or fall.
2. Acts as a reservoir that supplies the extra water when the turbine discharge is suddenly increased, and absorbs the water when the discharge is suddenly cut off.
3. Protects the tunnel from collapse and the penstock from over-pressure; reduces the length of the pipe in which the water hammer acts.
4. Improves speed regulation of the turbine (governor stability) by letting the tunnel flow change slowly.

Formulas (simple cylindrical tank, usual notations)

$L$ = tunnel length, $A_t$ = tunnel area, $A_s$ = surge tank area, $V_0$ = initial velocity in the tunnel, $g$ = gravity, $h_f$ = friction loss at $V_0$ ($h_f=\alpha V_0^2$), $H_0$ = gross head.

where $\alpha=\dfrac{h_f}{V_0^2}$. For a partial change of discharge, $V_0$ is replaced by the change $\Delta V$ in the tunnel velocity. The area of the tank is taken as 1.0-1.5 times $A_{th}$ (usually larger when the upsurge/downsurge governs the design).

Part 1: Derivation (no friction)

Let $z$ be the water level in the tank above the reservoir level (positive upward), $V$ the velocity in the tunnel (length $L$, area $A_t$), $A_s$ the area of the tank, and $Q_t$ the turbine flow after the sudden change.

 reservoir ~~~~~~~~
            \___________ tunnel L, A_t, V ___________ surge tank
                                              | z    |  A_s
                                              |______|__ to turbine Q_t

Continuity at the tank junction: inflow from the tunnel = flow to the tank + flow to the turbine

$$A_tV=A_s\frac{dz}{dt}+Q_t \quad (1)$$

Momentum for the water column in the tunnel (net force = mass x acceleration; the head $z$ acts to decelerate the flow):

$$\rho A_tL\frac{dV}{dt}=-\rho gA_t z\ \Rightarrow\ \frac{L}{g}\frac{dV}{dt}=-z \quad (2)$$

Differentiate (1) with $Q_t$ constant: $A_t\dfrac{dV}{dt}=A_s\dfrac{d^2z}{dt^2}$. Substituting in (2):

$$\frac{d^2z}{dt^2}+\frac{gA_t}{LA_s}\,z=0$$

This is simple harmonic motion with $\omega=\sqrt{\dfrac{gA_t}{LA_s}}$, so the period is $T=2\pi\sqrt{\dfrac{LA_s}{gA_t}}$.

Boundary conditions at $t=0$ (sudden change of discharge from $Q_0$ to $Q_t$, so $\Delta V=\dfrac{Q_0-Q_t}{A_t}$): $z=0$ and $\dfrac{dz}{dt}=\dfrac{A_t\Delta V}{A_s}$. Hence

$$z=\frac{A_t\Delta V}{A_s\,\omega}\sin\omega t,\qquad Z_{max}=\frac{A_t\Delta V}{A_s}\sqrt{\frac{LA_s}{gA_t}}$$ $$\boxed{Z_{max}=\Delta V\sqrt{\frac{L\,A_t}{g\,A_s}}}$$

For load rejection this is the upsurge $+Z_{max}$; for sudden demand the same magnitude is the downsurge $-Z_{max}$ (undamped).

Part 2: Numerical design

Given: reservoir RL 320 m, tailwater RL 200 m, gross head $H_0=120$ m; $Q=100$ m³/s for three units, so each unit takes $33.3$ m³/s; two units are rejected or demanded, so the change is $\Delta Q=66.7$ m³/s. Maximum and minimum tank levels are RL 330 m and RL 310 m, so $Z=330-320=10$ m upwards and $320-310=10$ m downwards. Tunnel $D=7$ m, tunnel head loss $h_f=10\%H_0=12$ m. Penstocks: 3 x 500 m, $D=2.5$ m, $f=0.016$. The friction factor of the tunnel is assumed to be the same as the penstocks, $f=0.016$.

Tunnel velocity and change in velocity

$$A_t=\frac{\pi}{4}(7)^2=38.48\ \text{m}^2,\quad V_0=\frac{100}{38.48}=2.598\ \text{m/s},\quad \Delta V=\frac{2}{3}V_0=1.732\ \text{m/s}$$

Length of the tunnel (from the head loss, Darcy-Weisbach)

$$h_f=\frac{f\,L}{D}\frac{V_0^2}{2g}\ \Rightarrow\ L=\frac{2gD\,h_f}{fV_0^2}=\frac{19.62\times7\times12}{0.016\times6.752}=15256\ \text{m}\ (\approx15.3\ \text{km})$$

Area of the tank from the permissible surge ($Z=10$ m both ways)

$$Z^2=\Delta V^2\frac{L\,A_t}{g\,A_s}\ \Rightarrow\ A_s=\frac{\Delta V^2\,L\,A_t}{g\,Z^2}=\frac{3.001\times15256\times38.48}{9.81\times100}=1796\ \text{m}^2$$

This corresponds to a circular tank of diameter $\sqrt{4A_s/\pi}=47.8$ m.

Thoma check (stability)

$$\alpha=\frac{h_f}{V_0^2}=\frac{12}{6.752}=1.777\ \text{s}^2/\text{m}$$

Penstock velocity $V_p=\dfrac{33.3}{\frac{\pi}{4}(2.5)^2}=6.79$ m/s and its loss $h_p=\dfrac{0.016\times500}{2.5}\dfrac{6.79^2}{19.62}=7.52$ m. Net head $H_0-h_f-h_p=120-12-7.52=100.5$ m.

$$A_{th}=\frac{L\,A_t}{2g\,\alpha\,(H_0-h_f-h_p)}=\frac{15256\times38.48}{19.62\times1.777\times100.5}=168\ \text{m}^2$$

Since $A_s=1796$ m² $>A_{th}=168$ m², the tank is stable.

Period of oscillation: $T=2\pi\sqrt{\dfrac{LA_s}{gA_t}}=1693$ s (about 28.2 min).

Answer: minimum area of the surge tank $A_s\approx1800$ m² (diameter about 48 m) and the permissible tunnel length $L\approx15250$ m (about 15.3 km), with the surge limited to RL 330 m and RL 310 m.

In a simple cylindrical surge tank the tank is open to the tunnel through a full-size connection, so the tank level responds freely and the oscillations decay slowly, with high upsurge and deep downsurge. In a restricted orifice tank, the connection to the tunnel is narrowed by an orifice of small area.

Why the restricted orifice type is more efficient:

• Damping: when the water flows in or out of the tank, the orifice creates an extra head loss $h_o=\dfrac{Q^2}{2gC_d^2A_o^2}$. This opposes the motion, so the oscillation dies out sooner and the amplitude of the maximum upsurge and minimum downsurge is smaller.
• Quick action: on a load change, the orifice head loss at once builds a head difference between the tunnel and the tank, so the tunnel flow is retarded sooner and less water has to be stored or supplied. The tank can therefore be smaller (less volume and cost) for the same permissible surge.
• Better stability of the governed system, thanks to the greater damping.

A too-small orifice produces a high water-hammer pressure in the tunnel (it behaves almost like a closed valve), so the area is selected as a compromise (typically about 0.25-0.5 of the tunnel area, giving a head loss of the order of the surge height).

Given and tunnel length

$A_t = \frac{\pi}{4}(5)^2 = 19.635\ \text{m}^2$, $V = 28/19.635 = 1.426$ m/s, $A_s = 300\ \text{m}^2$, $h_f = 0.025 \times 200 = 5.0$ m.

$$L = \frac{h_f\,2gD}{fV^2} = \frac{5.0 \times 2(9.81)(5)}{0.015 \times 1.426^2} = 16{,}080\ \text{m}$$

Total tunnel length $\approx 16.1$ km.

Maximum upsurge and downsurge

$Z_0 = 1.426\sqrt{\dfrac{16080 \times 19.635}{9.81 \times 300}} = 14.77$ m (no friction).

With friction ($h_f = 5$ m) the upsurge equation gives $Z_{max} = 11.64$ m above reservoir level. The first downswing after the upsurge reaches $8.57$ m below reservoir level.

Factor of safety (Thoma stability)

Head-loss coefficient $\alpha = h_f/V^2 = 5/1.426^2 = 2.459\ \text{s}^2/\text{m}$. Net head $H - h_f = 195$ m.

$$A_{th} = \frac{L A_t}{2g\alpha (H-h_f)} = \frac{16080 \times 19.635}{2(9.81)(2.459)(195)} = 33.56\ \text{m}^2$$ $$\text{FOS} = \frac{A_s}{A_{th}} = \frac{300}{33.56} = 8.94$$

Answer: $L \approx 16{,}080$ m; upsurge $= 11.64$ m; downsurge $= 8.57$ m; factor of safety $= 8.94$ (tank is very stable, since $A_s \gg A_{th}$).