Hydro-electric Machines

Power

$$P = \eta \gamma Q H = 0.82 \times 9.81 \times 25 \times 150 = 30{,}166\ \text{kW}$$

(With $\eta = 0.81$: $P = 29{,}798$ kW.) One unit is adopted.

Specific speed

For Francis turbines, an empirical (Schweiger and Gregory) relation in metric units (kW, m, rpm) is

$$N_s = \frac{3763}{H^{0.854}} + 49.4 = \frac{3763}{150^{0.854}} + 49.4 = 101.5$$

Speed from $N_s = \dfrac{N\sqrt{P}}{H^{5/4}}$:

$$N = \frac{101.5 \times 150^{1.25}}{\sqrt{30166}} = 306.9\ \text{rpm}$$

Synchronous speed for 50 Hz is $N = 6000/p$ rpm. Nearest value: $p = 20$ poles, $N = 300$ rpm. Actual specific speed $N_s = \dfrac{300\sqrt{30166}}{150^{1.25}} = 99.3$ (98.7 for $\eta=0.81$).

Runner diameter

Peripheral velocity $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (typical 0.6 to 0.9 for this $N_s$):

$$u = 0.7\sqrt{2 \times 9.81 \times 150} = 37.97\ \text{m/s}$$ $$D = \frac{60u}{\pi N} = \frac{60 \times 37.97}{\pi \times 300} = 2.42\ \text{m}$$

Setting (Thoma cavitation coefficient)

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0432(0.993)^2 = 0.0426$$

Assume sea-level site: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0426 \times 150 = 3.68\ \text{m}$$

This is the highest permissible position of the runner outlet above tailwater level. A lower setting (a safety margin of 1 to 2 m, or lower for a higher site elevation) is adopted in practice. At higher elevation reduce $H_{atm}$ by about 1 m per 900 m.

Answer: $N_s \approx 99$ (target 101.5); $N = 300$ rpm; $D \approx 2.42$ m; $H_s \le 3.7$ m above tailwater level (practically set lower).

Cavitation is the formation and collapse of vapour bubbles in flowing water where local pressure falls to the vapour pressure. In reaction turbines it occurs mainly at the runner outlet (back of the blades) and in the draft tube inlet, where pressure is lowest.

Effects

• Pitting and erosion of runner blades, draft tube cone and guide vanes by the high-pressure impact of collapsing bubbles.
• Loss of efficiency and power because flow passages are blocked by vapour and the blade profile is damaged.
• Noise and vibration, which can damage bearings and cause fatigue failure.
• Reduced life, with frequent repair (welding and grinding) and plant outage.
• Unstable operation and surging in the draft tube at part load.

Preventive measures

1. Set the runner low enough: keep the turbine setting $H_s \le H_{atm} - H_v - \sigma H$, i.e. closer to or below tailwater level.
2. Use a suitable specific speed; lower $N_s$ (a smaller $\sigma_c$) for high heads.
3. Use a well-designed draft tube with a gradual divergence (included angle 8 to 10 degrees) to recover pressure.
4. Use cavitation-resistant materials: stainless steel (13Cr-4Ni) runners, stellite or stainless overlay welding.
5. Maintain smooth, polished blade surfaces and correct blade profile, avoid sharp edges.
6. Avoid operation at very low part load or overload; admit air through the draft tube or runner centre (aeration) to cushion bubble collapse.
7. Keep silt out of the water (silt makes erosion worse), and carry out regular inspection and repair.

Purpose of a draft tube

A draft tube is a gradually expanding pipe connecting the runner outlet to the tailrace. It is provided to:

• allow the turbine to be set above tailwater level (for easy access) without losing the head between runner and tailwater;
• regain kinetic energy at runner outlet by reducing velocity from $V_1$ to $V_2$ and converting it into pressure, thus creating a suction (low pressure) at the runner exit and increasing the effective head;
• discharge water safely into the tailrace.

   runner outlet (1)
      |  |            ^
      |  |            | H_s
 -----+  +----------  | 
       \  /  TWL ~~~~~|~~~~
        \/   (2)  exit under water

Maximum permissible setting

Let section 1 be the draft tube inlet (runner outlet), at height $H_s$ above tailwater, with velocity $V_1$ and pressure $p_1$. Section 2 is the draft tube outlet submerged in tailwater, velocity $V_2$, pressure $\approx p_{atm}$. Bernoulli's equation between 1 and 2 (datum at tailwater) with loss $h_f$ in the tube:

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} + H_s = \frac{p_{atm}}{\gamma} + \frac{V_2^2}{2g} + h_f$$ $$\frac{p_1}{\gamma} = H_{atm} - H_s - \left[\frac{V_1^2 - V_2^2}{2g} - h_f\right]$$

Draft tube efficiency

It is the ratio of the actual kinetic energy regained to the kinetic energy at inlet:

$$\eta_d = \frac{\dfrac{V_1^2 - V_2^2}{2g} - h_f}{\dfrac{V_1^2}{2g}}$$

Substituting in the pressure equation,

$$\frac{p_1}{\gamma} = H_{atm} - H_s - \eta_d\frac{V_1^2}{2g}$$

To avoid cavitation, $p_1 \ge p_v$ (vapour pressure), so

$$H_{s,max} = H_{atm} - H_v - \eta_d\frac{V_1^2}{2g}$$

In terms of the Thoma coefficient $\sigma = \dfrac{H_{atm} - H_v - H_s}{H}$:

$$H_{s,max} = H_{atm} - H_v - \sigma_c H$$

A smaller $\sigma_c$ allows a higher setting. The exit loss $V_2^2/2g$ is lost to the tailrace, so $\eta_d$ is typically 0.75 to 0.90.

The draft tube is the diverging pipe from the runner outlet to the tailrace. Its functions are:

1. Allows the turbine to be placed above tailwater (convenient, dry installation) without losing the head between runner exit and tailwater level.
2. Converts kinetic energy into pressure energy: the gradually increasing area reduces velocity, recovering most of the exit velocity head $V_1^2/2g$ that would otherwise be wasted.
3. Creates suction at the runner outlet, so the pressure there is below atmospheric and the effective head on the runner is increased.
4. Discharges water to the tailrace smoothly, submerged below tailwater, which keeps the tube full and sealed against air entry.
5. Reduces exit losses and increases the overall efficiency of the turbine, especially for low-head, high-speed turbines.

Sketch (vertical section through a vertical-shaft Francis unit)

  Penstock
     |
     v         generator
  +--+---+     ====|====
  | Valve|        shaft
  +--+---+         |
     |       stay ring/guide vanes
   ==+==============+====
  /   SPIRAL CASING   \   <- surrounds the runner
  \_____ runner _______/
          |   |
          |   |  DRAFT TUBE
          \   /   (diverging pipe)
           \ /
 TWL ~~~~~~~~V~~~~~~~~~~~~~~> tailrace

The spiral (scroll) casing surrounds the guide vanes and runner at the top; the draft tube is directly below the runner and ends under tailwater.

Importance

• Spiral casing: distributes water uniformly all around the guide vanes and runner with constant velocity (its area decreases along the flow path), so the flow enters the runner without shock and with low loss. It also withstands the water pressure and supports the turbine.
• Draft tube: recovers kinetic energy at the runner exit, allows turbine setting above tailwater, increases the effective head, and conveys water to the tailrace.

Turbines are classified by the action of water on the runner: impulse turbines (jet at atmospheric pressure) and reaction turbines (runner fully submerged, pressure drops across the runner).

Impulse turbines

• Pelton wheel: one to six jets strike double-cup buckets on a wheel. Suited to high heads (above about 300 m, down to about 100 m), low discharge, $N_s$ about 4 to 70. Mounted above tailwater.
• Turgo: inclined jet, runs at higher speed; medium heads (50 to 250 m).
• Cross-flow (Banki-Michell): low-cost, for small hydro, heads 5 to 100 m and small discharge.

Reaction turbines

• Francis: radial inlet, axial outlet, with spiral casing, guide vanes and draft tube. Medium heads (about 30 to 500 m), $N_s$ about 60 to 400. Most widely used.
• Kaplan: axial flow with adjustable runner blades and guide vanes (double regulated). Low heads (up to about 70 m), large discharges, $N_s$ about 300 to 1000.
• Propeller: like Kaplan with fixed blades; suited to nearly constant load.
• Bulb / Straflo: horizontal axial turbines with the generator in a bulb or at the rim; very low heads (below 20 m), run-of-river and tidal schemes.

Suitability and performance

Major performance characteristics

• Efficiency vs load: Pelton and Kaplan keep efficiency above about 85 % from 25 to 100 % load; Francis has a peak near 80 to 90 % load and falls at light load; propeller falls sharply.
• Specific speed: increases from Pelton to Francis to Kaplan, and determines the type for a given head.
• Runaway speed: about 1.8 to 1.9 $N$ for Pelton, 2 to 2.2 $N$ for Francis, 2.5 to 3 $N$ for Kaplan.
• Cavitation: a risk for reaction turbines, controlled by the Thoma coefficient and setting; absent in Pelton.
• Typical peak efficiency: 90 % for Pelton, 92 to 95 % for Francis and 92 to 94 % for Kaplan.

Efficiency curves

 Eff %
 100|
  90|        __Pelton___
  80|      /      ___Francis_
  70|     /     /        \
  60|    /    /            \
  50|   /   /
  40|  / /
    +----+----+----+----+---- % load
        25   50   75  100

Performance characteristics

• Pelton: efficiency is high (about 85 to 90 %) over a wide range of load (about 30 to 100 %), because load is regulated by changing the number of jets or nozzle spear position, so each jet works at its best speed ratio.
• Francis: peak efficiency (about 92 to 95 %) occurs near the design load; at part load the guide vane opening alters the flow angle at runner inlet, causing shock and a fall of efficiency below about 50 % load.
• Both curves are approximately parabolic in shape; head variation changes the best-efficiency speed, more strongly for Francis.

Advantage of Pelton over Francis

• Nearly flat efficiency curve at part load; suitable for plants with varying load or flow.
• No cavitation, no draft tube, easy to inspect and repair buckets.
• More tolerant of silt, since only buckets and nozzles need replacement and are easy to access.
• Best for high heads where Francis would be too fast or heavily stressed.

Principle of setting a Francis turbine

The runner is set so that the pressure at the runner exit (draft tube inlet) never falls below the vapour pressure, to avoid cavitation. The maximum height of the runner outlet above tailwater is

$$H_s = H_{atm} - H_v - \sigma_c H$$

where $\sigma_c$ is the Thoma cavitation coefficient (a function of specific speed). Higher $N_s$ means a larger $\sigma_c$, so the turbine must be set lower, and at high elevation (lower $H_{atm}$) it must also be set lower; the setting may be below tailwater level.

Francis turbines are preferred to Pelton turbines in the following conditions:

1. Medium head range: about 30 to 500 m, where the Francis has a higher specific speed and a smaller, faster runner. (Pelton is preferred above about 300 to 500 m.)
2. Large discharge: for a given power at medium head, a Francis handles a large flow in a compact runner; a Pelton would need many jets and a very large wheel.
3. High specific speed needed: a Francis runs faster, so it can be directly coupled to a smaller, cheaper generator.
4. Compactness and lower civil cost: smaller powerhouse, with the unit placed on a vertical shaft close to tailwater; lower cost per kW at moderate heads.
5. Utilisation of full head: the draft tube lets the turbine use the head between runner and tailwater, which matters when this height is a large part of the head (low and medium head). A Pelton wheel must stay above tailwater and loses this head.
6. Higher peak efficiency at design load (up to about 95 %) and a nearly constant load.
7. Clean water, steady load: where the sediment is small (otherwise runner erosion is severe) and load variation is small.
8. Foundation and weight: better where a lower, more rigid foundation is possible; avoids very high-pressure nozzles and long penstock.

In summary, choose Francis for medium head, large flow and steady load; choose Pelton for high head, low flow, silty water or widely varying load.

Setting of a turbine

Setting means fixing the elevation of the turbine (runner outlet or centre line) relative to the tailwater level. For a reaction turbine it is limited by cavitation, $H_s \le H_{atm} - H_v - \sigma_c H$. For a Pelton wheel the wheel centre is set above the maximum flood level in the tailrace (clearance 1 to 2 m) so that the runner never dips in water.

Data

Net head at nozzle $H = 360$ m, $C_v = 0.98$, $\eta = 0.85$, $N = 560$ rpm, $N_s = 15.3$ (for each single-jet wheel), $L = 1200$ m, $h_f = 18$ m, $f = 0.02$ (Darcy).

(i) Total power developed

Power of one wheel from specific speed (kW, m, rpm):

$$P = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{15.3 \times 360^{1.25}}{560}\right)^2 = 1835.5\ \text{kW}$$ $$P_{total} = 3 \times 1835.5 = 5506.6\ \text{kW}$$

(ii) Discharge

$$Q = \frac{P_{total}}{\eta \gamma H} = \frac{5506.6}{0.85 \times 9.81 \times 360} = 1.834\ \text{m}^3/\text{s}$$

Per wheel (one jet): $q = 0.6115\ \text{m}^3/\text{s}$.

(iii) Jet diameter and pipe diameter

Jet velocity $V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 360} = 82.36$ m/s.

$$d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 0.6115}{\pi \times 82.36}} = 0.0972\ \text{m}$$

Pipeline: $h_f = \dfrac{fLv^2}{2gD}$ with $v = \dfrac{4Q}{\pi D^2}$:

$$D^5 = \frac{16 f L Q^2}{\pi^2 \, 2g \, h_f} = \frac{16 \times 0.02 \times 1200 \times 1.834^2}{\pi^2 \times 2 \times 9.81 \times 18}$$ $$D = 0.820\ \text{m},\quad v = 3.47\ \text{m/s}$$

Answer: total power $= 5507$ kW; $Q = 1.834$ m³/s; jet diameter $= 97$ mm; pipe diameter $= 0.82$ m.

Data

$N = 190$ rpm, $P = 5150.25$ kW, $H = 220$ m, $\eta_o = 0.80$.

Discharge from power:

$$Q = \frac{P}{\eta_o \gamma H} = \frac{5150.25}{0.8 \times 9.81 \times 220} = 2.983\ \text{m}^3/\text{s}$$

i) Unit quantities (head 1 m)

$$N_u = \frac{N}{\sqrt{H}} = \frac{190}{\sqrt{220}} = 12.81\ \text{rpm}$$ $$Q_u = \frac{Q}{\sqrt{H}} = \frac{2.983}{\sqrt{220}} = 0.2011\ \text{m}^3/\text{s}$$ $$P_u = \frac{P}{H^{3/2}} = \frac{5150.25}{220^{1.5}} = 1.578\ \text{kW}$$

ii) At $H = 140$ m (same efficiency, similar operation)

$$N = N_u\sqrt{140} = 12.81 \times 11.832 = 151.6\ \text{rpm}$$ $$Q = Q_u\sqrt{140} = 0.2011 \times 11.832 = 2.380\ \text{m}^3/\text{s}$$ $$P = P_u \times 140^{1.5} = 1.578 \times 1656.5 = 2614\ \text{kW}$$

Answer: $N_u = 12.81$ rpm, $Q_u = 0.2011$ m³/s, $P_u = 1.578$ kW; at 140 m: $N = 151.6$ rpm, $Q = 2.38$ m³/s, $P = 2614$ kW.

Data

$P = 70$ kW (shaft), $H = 100$ m, $N = 400$ rpm, $\eta_o = 0.85$, speed ratio $\phi = u/V_1 = 0.46$, penstock 200 mm.

(i) Volumetric flow rate

$$Q = \frac{P}{\eta_o \gamma H} = \frac{70}{0.85 \times 9.81 \times 100} = 0.0839\ \text{m}^3/\text{s}$$

($= 83.9$ litres per second; velocity in the 200 mm penstock $= 2.67$ m/s.)

(ii) Wheel diameter

Jet velocity (taking $C_v \approx 1$, as no nozzle coefficient is given): $V_1 = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 100} = 44.29$ m/s.

Bucket speed $u = 0.46 \times 44.29 = 20.37$ m/s.

$$D = \frac{60u}{\pi N} = \frac{60 \times 20.37}{\pi \times 400} = 0.972\ \text{m}$$

(If $C_v = 0.98$ were used, $V_1 = 43.40$ m/s, $u = 19.97$ m/s and $D = 0.953$ m.)

Answer: $Q = 0.0839$ m³/s; wheel diameter $\approx 0.97$ m.

Data

$H = 200$ m, $Q = 4\ \text{m}^3/\text{s}$, $C_v = 0.98$, $D = 1.47$ m, $\eta_m = 0.86$, $\phi = 0.47$, $d/D = 0.113$.

Jet velocity: $V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 200} = 61.39$ m/s.

(i) Rotational speed

$u = \phi V_1 = 0.47 \times 61.39 = 28.85$ m/s

$$N = \frac{60u}{\pi D} = \frac{60 \times 28.85}{\pi \times 1.47} = 374.9\ \text{rpm}$$

(ii) Power output

Taking the 86 % efficiency as the efficiency from the water power to the shaft:

$$P = \eta\,\gamma Q H = 0.86 \times 9.81 \times 4 \times 200 = 6749\ \text{kW} = 6.75\ \text{MW}$$

(iii) Number of nozzles

Jet diameter $d = 0.113 \times 1.47 = 0.1661$ m; jet area $= 0.02167\ \text{m}^2$.

Discharge per jet $q = 0.02167 \times 61.39 = 1.330\ \text{m}^3/\text{s}$.

$$n = \frac{Q}{q} = \frac{4}{1.330} = 3.01 \approx 3\ \text{nozzles}$$

(iv) Specific speed

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{374.9\sqrt{6749}}{200^{1.25}} = 40.9$$

Answer: $N = 374.9$ rpm; $P = 6.75$ MW; 3 nozzles; $N_s = 40.9$ (kW, m, rpm).

Step 1: Power

$$P = \eta\gamma Q H = 0.85 \times 9.81 \times 5 \times 312.5 = 13{,}029\ \text{kW}$$

Step 2: Jet velocity and bucket speed

$V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 312.5} = 76.74$ m/s

Speed ratio $K_u = u/\sqrt{2gH} = 0.46$ (usual range 0.43 to 0.47) with $\sqrt{2gH} = 78.30$ m/s: $u = 0.46 \times 78.30 = 36.02$ m/s

Step 3: Number of jets and speed

For a single jet, the empirical (Siervo and Lugaresi) specific speed is $N_s \approx 85.49/H^{0.243} = 21.2$. For $z$ jets the wheel specific speed is $N_s\sqrt{z}$. Take $z = 2$ jets (horizontal shaft):

$$N = \frac{N_s\sqrt{z}\,H^{5/4}}{\sqrt{P}} = 344.6\ \text{rpm}$$

Synchronous speed for 50 Hz: $N = 6000/p$. Choose $p = 18$ poles, $N = 333.3$ rpm.

Step 4: Wheel (pitch circle) diameter

$$D = \frac{60u}{\pi N} = \frac{60 \times 36.02}{\pi \times 333.3} = 2.064\ \text{m}$$

Step 5: Jet diameter

Discharge per jet $q = 5/2 = 2.5\ \text{m}^3/\text{s}$:

$$d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 2.5}{\pi \times 76.74}} = 0.2037\ \text{m}$$

Jet ratio $m = D/d = 10.13$ (acceptable range 10 to 20).

Step 6: Bucket details

• Number of buckets $Z = 15 + \dfrac{D}{2d} = 15 + 5.07 \approx 20$
• Bucket width $B \approx 3.5d = 0.71$ m; depth about $1.0d$; length about $2.5d$.

Step 7: Specific speed of the wheel

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{333.3\sqrt{13{,}029}}{312.5^{1.25}} = 29.0$$

(about 20.5 per jet)

Answer: $P = 13.03$ MW; 2 jets; $N = 333.3$ rpm; $D = 2.06$ m; $d = 0.204$ m ($D/d = 10.1$); about 20 buckets; $N_s = 29.0$.

Step 1: Power

$$P = \eta\gamma Q H = 0.90 \times 9.81 \times 5 \times 310 = 13{,}685\ \text{kW}$$

Step 2: Jet velocity and bucket speed

$V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 310} = 76.43$ m/s

Speed ratio $K_u = u/\sqrt{2gH} = 0.46$ (usual range 0.43 to 0.47) with $\sqrt{2gH} = 77.99$ m/s: $u = 0.46 \times 77.99 = 35.87$ m/s

Step 3: Number of jets and speed

For a single jet, the empirical (Siervo and Lugaresi) specific speed is $N_s \approx 85.49/H^{0.243} = 21.2$. For $z$ jets the wheel specific speed is $N_s\sqrt{z}$. Take $z = 2$ jets (horizontal shaft; 50 Hz assumed):

$$N = \frac{N_s\sqrt{z}\,H^{5/4}}{\sqrt{P}} = 333.5\ \text{rpm}$$

Synchronous speed for 50 Hz: $N = 6000/p$. Choose $p = 18$ poles, $N = 333.3$ rpm.

Step 4: Wheel (pitch circle) diameter

$$D = \frac{60u}{\pi N} = \frac{60 \times 35.87}{\pi \times 333.3} = 2.056\ \text{m}$$

Step 5: Jet diameter

Discharge per jet $q = 5/2 = 2.5\ \text{m}^3/\text{s}$:

$$d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 2.5}{\pi \times 76.43}} = 0.2041\ \text{m}$$

Jet ratio $m = D/d = 10.07$ (acceptable range 10 to 20).

Step 6: Bucket details

• Number of buckets $Z = 15 + \dfrac{D}{2d} = 15 + 5.04 \approx 20$
• Bucket width $B \approx 3.5d = 0.71$ m; depth about $1.0d$; length about $2.5d$.

Step 7: Specific speed of the wheel

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{333.3\sqrt{13{,}685}}{310^{1.25}} = 30.0$$

(about 21.2 per jet)

Answer: $P = 13.68$ MW; 2 jets; $N = 333.3$ rpm; $D = 2.06$ m; $d = 0.204$ m; specific speed of the turbine $N_s \approx 30$ (kW, m, rpm), i.e. 21 per jet.

Data

$P = 6867$ kW, $H = 350$ m, $\eta_o = 0.80$, $N = 550$ rpm, $K_v = 0.98$, $K_u = 0.46$, $d/D = 1/12$.

Discharge

$$Q = \frac{P}{\eta_o \gamma H} = \frac{6867}{0.8 \times 9.81 \times 350} = 2.50\ \text{m}^3/\text{s}$$

Wheel diameter

$V_1 = K_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 350} = 81.21$ m/s

$u = K_u\sqrt{2gH} = 0.46 \times 82.87 = 38.12$ m/s

$$D = \frac{60u}{\pi N} = \frac{60 \times 38.12}{\pi \times 550} = 1.324\ \text{m}$$

Jet diameter and number of jets

From $d/D = 1/12$: $d = 1.324/12 = 0.1103$ m.

Discharge through one jet $= \dfrac{\pi}{4}(0.1103)^2 \times 81.21 = 0.776\ \text{m}^3/\text{s}$.

$$z = \frac{2.50}{0.776} = 3.22 \Rightarrow 4\ \text{jets}$$

With 4 jets the jet diameter is adjusted to carry the full discharge:

$$d = \sqrt{\frac{4 \times 2.5/4}{\pi \times 81.21}} = 0.099\ \text{m} \approx 99\ \text{mm}$$

so $D/d = 13.4$ (within the usual range 10 to 20). Specific speed: $N_s = \dfrac{550\sqrt{6867}}{350^{1.25}} = 30.1$.

Answer: $Q = 2.50$ m³/s; 4 jets; jet diameter $\approx 0.099$ m; wheel diameter $= 1.32$ m.

Interpretation

The 100 mm diameter is taken as the diameter of each jet (nozzle outlet). Head $H = 300$ m, $C_v = 0.96$, total $Q = 1.5\ \text{m}^3/\text{s}$.

Velocity of jet

$$V_1 = C_v\sqrt{2gH} = 0.96\sqrt{2 \times 9.81 \times 300} = 73.65\ \text{m/s}$$

Discharge through one jet

$$q = \frac{\pi}{4}d^2V_1 = \frac{\pi}{4}(0.1)^2 \times 73.65 = 0.5785\ \text{m}^3/\text{s}$$

Number of jets

$$n = \frac{Q}{q} = \frac{1.5}{0.5785} = 2.59 \approx 3$$

Answer: 3 jets (the next whole number above 2.59).

Step 1: Power and units

$$P = \eta\gamma Q H = 0.86 \times 9.81 \times 40 \times 90 = 30{,}372\ \text{kW}$$

One unit is adopted ($P_u = 30{,}372$ kW).

Step 2: Specific speed and rated speed

For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is

$$N_s = \frac{3763}{H^{0.854}} + 49.4 = 130.1$$ $$N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 206.9\ \text{rpm}$$

Synchronous speed at 50 Hz: $N = 6000/p$. Adopt $p = 30$ poles, $N = 200$ rpm. Actual specific speed:

$$N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 125.7$$

Step 3: Runner diameter

Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9):

$$u = 0.7\sqrt{2 \times 9.81 \times 90} = 29.41\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 29.41}{\pi \times 200} = 2.81\ \text{m}$$

Step 4: Setting (Thoma cavitation coefficient)

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0683$$

Sea-level site assumed: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0683 \times 90 = 3.91\ \text{m}$$

This is the highest permissible level of the runner outlet above tailwater; in practice set lower (safety margin 1 to 2 m, and reduce $H_{atm}$ by about 1 m per 900 m of site elevation).

Answer: $P = 30.4$ MW; specific speed $N_s \approx 126$ (design value 130); $N = 200$ rpm; runner diameter $D \approx 2.81$ m; setting $H_s \le 3.9$ m above tailwater level.

Step 1: Power and units

$$P = \eta\gamma Q H = 0.94 \times 9.81 \times 60 \times 90 = 49{,}796\ \text{kW}$$

Number of units: a single 50 MW unit would need a large runner (3.6 m) and shaft. Adopt 2 units, each $Q = 30$ m³/s and $P_u = 24{,}898$ kW (gives easier transport, maintenance and part-load operation).

Step 2: Specific speed and rated speed

For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is

$$N_s = \frac{3763}{H^{0.854}} + 49.4 = 130.1$$ $$N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 228.5\ \text{rpm}$$

Synchronous speed at 50 Hz: $N = 6000/p$. Adopt $p = 26$ poles, $N = 230.8$ rpm. Actual specific speed:

$$N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 131.4$$

Step 3: Runner diameter

Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9):

$$u = 0.7\sqrt{2 \times 9.81 \times 90} = 29.41\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 29.41}{\pi \times 230.8} = 2.43\ \text{m}$$

Step 4: Setting (Thoma cavitation coefficient)

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0745$$

Sea-level site assumed: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0745 \times 90 = 3.35\ \text{m}$$

This is the highest permissible level of the runner outlet above tailwater; in practice set lower (safety margin 1 to 2 m, and reduce $H_{atm}$ by about 1 m per 900 m of site elevation).

Answer: 2 units of 24.9 MW; $N_s \approx 131$; $N = 230.8$ rpm; $D \approx 2.43$ m; setting $H_s \le 3.35$ m above tailwater level.

Step 1: Power and units

$$P = \eta\gamma Q H = 0.90 \times 9.81 \times 140 \times 110 = 135{,}967\ \text{kW}$$

Number of units: adopt 4 units, each $Q = 35$ m³/s and $P_u = 33{,}992$ kW.

Step 2: Specific speed and rated speed

For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is

$$N_s = \frac{3763}{H^{0.854}} + 49.4 = 117.3$$ $$N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 226.7\ \text{rpm}$$

Synchronous speed at 50 Hz: $N = 6000/p$. Adopt $p = 26$ poles, $N = 230.8$ rpm. Actual specific speed:

$$N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 119.4$$

Step 3: Runner diameter

Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9):

$$u = 0.7\sqrt{2 \times 9.81 \times 110} = 32.52\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 32.52}{\pi \times 230.8} = 2.69\ \text{m}$$

Step 4: Setting (Thoma cavitation coefficient)

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0616$$

Sea-level site assumed: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0616 \times 110 = 3.28\ \text{m}$$

This is the highest permissible level of the runner outlet above tailwater; in practice set lower (safety margin 1 to 2 m, and reduce $H_{atm}$ by about 1 m per 900 m of site elevation).

Answer: 4 units of 34 MW; $N = 230.8$ rpm (50 Hz, 26 poles); runner diameter $D \approx 2.69$ m; runner outlet at about 3.3 m above tailrace water level as the upper limit (set lower in practice).

Step 1: Power and units

$$P = \eta\gamma Q H = 0.85 \times 9.81 \times 160 \times 150 = 200{,}124\ \text{kW}$$

Number of units: adopt 4 units, each $Q = 40$ m³/s and $P_u = 50{,}031$ kW.

Step 2: Specific speed and rated speed

For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is

$$N_s = \frac{3763}{H^{0.854}} + 49.4 = 101.5$$ $$N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 238.3\ \text{rpm}$$

Synchronous speed at 50 Hz: $N = 6000/p$. Adopt $p = 26$ poles, $N = 230.8$ rpm. Actual specific speed:

$$N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 98.3$$

Step 3: Runner diameter

Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9):

$$u = 0.7\sqrt{2 \times 9.81 \times 150} = 37.97\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 37.97}{\pi \times 230.8} = 3.14\ \text{m}$$

Step 4: Setting (Thoma cavitation coefficient)

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0418$$

Sea-level site assumed: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0418 \times 150 = 3.79\ \text{m}$$

This is the highest permissible level of the runner outlet above tailwater; in practice set lower (safety margin 1 to 2 m, and reduce $H_{atm}$ by about 1 m per 900 m of site elevation).

Answer: 4 units of 50 MW; $N = 230.8$ rpm; runner diameter $D \approx 3.14$ m; setting height $H_s \le 3.8$ m above tailwater level.

Given

$Q = 23\ \text{m}^3/\text{s}$, $H = 20$ m, $\eta = 0.85$, $N = 170$ rpm, $N_s = 230$, $K_u = 0.76$.

Total power available

$$P_{total} = \eta\gamma Q H = 0.85 \times 9.81 \times 23 \times 20 = 3836\ \text{kW}$$

Power per turbine from specific speed

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} \;\Rightarrow\; P = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{230 \times 20^{1.25}}{170}\right)^2 = 3274\ \text{kW}$$

Number of turbines

$$n = \frac{3836}{3274} = 1.17 \Rightarrow 2\ \text{turbines}$$

(One turbine of 3274 kW cannot carry the whole 3836 kW, so the next whole number is used. Each then develops $3836/2 = 1918$ kW and takes $Q = 11.5\ \text{m}^3/\text{s}$.)

Runner diameter

$$u = K_u\sqrt{2gH} = 0.76\sqrt{2 \times 9.81 \times 20} = 15.05\ \text{m/s}$$ $$D = \frac{60u}{\pi N} = \frac{60 \times 15.05}{\pi \times 170} = 1.69\ \text{m}$$

Answer: 2 turbines; runner diameter $\approx 1.69$ m.

Data

$P = 32$ MW, $H = 54$ m, draft tube inlet $d_1 = 3$ m, inlet at $H_s = 2.2$ m above tailrace level, vacuum at inlet $= 4.8$ m of water, $\eta_d = 0.78$.

Energy regained in the draft tube

Apply Bernoulli between the inlet (1) and the tailrace surface (2), using gauge pressures ($p_1/\gamma = -4.8$ m, tailrace surface 0):

$$-4.8 + \frac{V_1^2}{2g} + 2.2 = \frac{V_2^2}{2g} + h_f$$ $$\frac{V_1^2 - V_2^2}{2g} - h_f = 4.8 - 2.2 = 2.6\ \text{m}$$

By definition of draft tube efficiency,

$$\eta_d = \frac{\frac{V_1^2 - V_2^2}{2g} - h_f}{V_1^2/2g} \;\Rightarrow\; \frac{V_1^2}{2g} = \frac{2.6}{0.78} = 3.333\ \text{m}$$ $$V_1 = \sqrt{2 \times 9.81 \times 3.333} = 8.087\ \text{m/s}$$

Discharge and overall efficiency

$$Q = \frac{\pi}{4}(3)^2 \times 8.087 = 57.16\ \text{m}^3/\text{s}$$

Water power $= \gamma Q H = 9.81 \times 57.16 \times 54 = 30{,}282$ kW.

$$\eta_o = \frac{P}{\gamma Q H} = \frac{32{,}000}{30{,}282} = 1.057$$

Answer: the data as printed give $Q = 57.2\ \text{m}^3/\text{s}$ and $\eta_o = 1.057$ (105.7 %). This is above 100 %, which is physically impossible, so one of the given values (32 MW, 4.8 m vacuum or 2.2 m setting) is inconsistent. The method above is the required one: find $V_1$ from the draft tube efficiency and the vacuum reading, then $Q$, then $\eta_o = P/\gamma QH$. (For example, if the inlet were 2.2 m below tailwater instead, the regained head would be $4.8 + 2.2 = 7.0$ m, giving $Q = 93.8\ \text{m}^3/\text{s}$ and $\eta_o = 64.4$ %.)

Data

$H = 25$ m, $P = 11{,}760$ kW, $N = 120$ rpm, $H_{atm} = 10$ m, $H_v = 0.20$ m.

Specific speed

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{120\sqrt{11760}}{25^{1.25}} = 232.8$$

Thoma cavitation coefficient

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0432 \times (2.328)^2 = 0.2341$$

Maximum height (setting) of the draft tube

The maximum height of the runner outlet / draft tube inlet above the tailwater level is

$$H_s = H_{atm} - H_v - \sigma_c H = 10 - 0.20 - 0.2341 \times 25 = 3.95\ \text{m}$$

Answer: the straight draft tube may be set at most about 3.95 m above the tailwater level (a lower setting is safer).

Data

$H = 40$ m, $Q = 10\ \text{m}^3/\text{s}$, $N = 300$ rpm, $K_u = 0.85$, $K_f = 0.3$, $\eta_o = 0.80$, $\eta_h = 0.90$. Outlet radial: $V_{w2} = 0$, $V_{f2} = V_{f1}$.

$\sqrt{2gH} = \sqrt{2 \times 9.81 \times 40} = 28.01$ m/s.

a) Power developed

$$P = \eta_o\gamma Q H = 0.8 \times 9.81 \times 10 \times 40 = 3139\ \text{kW}$$

b) Inlet diameter and width

$u_1 = K_u\sqrt{2gH} = 0.85 \times 28.01 = 23.81$ m/s

$$D_1 = \frac{60u_1}{\pi N} = \frac{60 \times 23.81}{\pi \times 300} = 1.516\ \text{m}$$

$V_{f1} = K_f\sqrt{2gH} = 0.3 \times 28.01 = 8.40$ m/s

$$B_1 = \frac{Q}{\pi D_1 V_{f1}} = \frac{10}{\pi \times 1.516 \times 8.40} = 0.25\ \text{m}$$

(vane thickness neglected)

c) Guide vane angle at inlet

From $\eta_h = \dfrac{V_{w1}u_1}{gH}$:

$$V_{w1} = \frac{0.9 \times 9.81 \times 40}{23.81} = 14.83\ \text{m/s}$$ $$\tan\alpha = \frac{V_{f1}}{V_{w1}} = \frac{8.40}{14.83} \Rightarrow \alpha = 29.5^\circ$$

Runner vane angle at inlet: since $V_{w1} < u_1$, $\tan\theta = \dfrac{V_{f1}}{u_1 - V_{w1}} = \dfrac{8.40}{8.98}$, so $\theta = 43.1^\circ$ (acute vane angle with the wheel tangent).

d) Specific speed

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{300\sqrt{3139}}{40^{1.25}} = 167.1$$

e) Outlet diameter

Constant flow velocity $V_{f2} = 8.40$ m/s and radial (axial-direction) discharge, so the outlet area is a full circle:

$$\frac{\pi}{4}D_2^2 = \frac{Q}{V_{f2}} \Rightarrow D_2 = \sqrt{\frac{4 \times 10}{\pi \times 8.40}} = 1.23\ \text{m}$$

Powerhouse dimensions for three units (assumed thumb rules)

Each unit: 3.14 MW, $D_1 = 1.52$ m. Assume vertical shaft, spiral casing outer diameter about $2.7D_1 \approx 4.1$ m.

  PLAN (3 units + erection bay)        Section
 +--------+--------+--------+-----+      roof
 | unit 1 | unit 2 | unit 3 | EB  |    ==========  crane rail
 |  (G)   |  (G)   |  (G)   |     |     |  G  |  generator floor
 +--------+--------+--------+-----+     | TURB|  turbine floor
  25.5 m x 13 m                          \DT /  draft tube

Answer: $P = 3139$ kW; $D_1 = 1.52$ m, $B_1 = 0.25$ m; $\alpha = 29.5^\circ$; $N_s = 167.1$; $D_2 = 1.23$ m; powerhouse about 25.5 m x 13 m x 19 m.

Data

$Q = 300\ \text{m}^3/\text{s}$, $H = 30$ m, $N = 166.7$ rpm, $\eta_o = 0.88$.

Total power

$$P_{total} = \eta\gamma Q H = 0.88 \times 9.81 \times 300 \times 30 = 77{,}695\ \text{kW}$$

(i) Francis, $N_s \le 267$

Maximum power per unit:

$$P_u = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{267 \times 30^{1.25}}{166.7}\right)^2 = 12{,}646\ \text{kW}$$ $$n = \frac{77{,}695}{12{,}646} = 6.14 \Rightarrow 7\ \text{units}$$

(each 11,099 kW, $N_s = 250$)

(ii) Kaplan, $N_s \le 650$

$$P_u = \left(\frac{650 \times 30^{1.25}}{166.7}\right)^2 = 74{,}948\ \text{kW}$$ $$n = \frac{77{,}695}{74{,}948} = 1.04 \Rightarrow 2\ \text{units}$$

(each 38,848 kW, $N_s = 468$)

Answer: 7 Francis units, or 2 Kaplan units.