Hydro-Electric Machines

Governing of a hydraulic turbine is the automatic regulation of the flow of water to the turbine (by guide vanes, nozzle spear or runner blades) so that its speed remains constant when the load on the generator changes.

Why governing is necessary

1. Constant frequency: the generator is coupled directly to the turbine, and frequency $f = \dfrac{PN}{120}$. Electrical loads (motors, clocks, electronics) need constant frequency (50 Hz in Nepal), so the speed must stay at synchronous speed.
2. Load changes all the time: consumer demand varies through the day. If the load falls and water input stays the same, the extra energy accelerates the runner; if the load increases, the turbine slows down. Water input must follow the load.
3. Prevent runaway: on sudden full-load rejection (e.g. a line trip), an ungoverned turbine speeds up to runaway speed (1.8–2.5 times normal), causing dangerous centrifugal stresses and damage to generator and runner.
4. Constant voltage and stable parallel operation: speed variation affects voltage and the synchronism of machines running in parallel on the grid; governors share load among units in proportion to their droop settings.
5. Economy of water: water is used only in proportion to the power demanded, which saves stored water.
6. Safety against water hammer: governors close guide vanes or nozzles at a controlled rate, and use deflectors/relief valves, so that pressure rise in the penstock stays within limits.

 load drops -> speed rises -> governor senses ->
 closes guide vanes/spear -> less Q -> speed normal

A centrifugal pump is a rotodynamic pump that raises the pressure of a liquid by the centrifugal action of a rotating impeller: it converts mechanical energy into pressure energy through forced-vortex motion.

Main parts

Impeller (with backward-curved vanes), volute (spiral) casing, suction pipe with foot valve and strainer, delivery pipe with delivery valve, and the shaft driven by a motor.

              delivery pipe
                   ||
            _______||______
          /   volute casing \
         |     _______       |
         |    / \ | / \      |
  suction|===|-- (eye) --|   |
   pipe  |    \ / | \ /      |
         |     impeller      |
          \_________________/
   |
  foot valve + strainer (in sump)

Working

1. Priming: the pump casing and suction pipe are first completely filled with liquid (priming) to remove air; an impeller rotating in air cannot create enough suction.
2. Starting: the motor rotates the impeller with the delivery valve closed (minimum power at zero flow).
3. Forced vortex and suction: as the impeller rotates, the liquid in it rotates as a forced vortex. Centrifugal force throws the liquid radially outward from the eye towards the tip, creating a low pressure (partial vacuum) at the eye. Atmospheric pressure on the sump surface pushes liquid up the suction pipe into the eye.
4. Energy transfer: the vanes give the liquid high velocity and pressure. The rise in pressure head due to the forced vortex is $\dfrac{u_2^2 - u_1^2}{2g}$, where $u$ is the peripheral speed.
5. Conversion in casing: the liquid leaves the impeller with high kinetic energy and enters the volute casing, whose area increases gradually; the velocity is reduced and kinetic energy is converted into pressure energy.
6. Delivery: the delivery valve is opened and the high-pressure liquid flows up the delivery pipe to the required height.
7. Stopping: the delivery valve is closed before stopping the motor to avoid backflow and water hammer.

A centrifugal pump works only after priming and is used for large discharge at moderate heads (water supply, irrigation, dewatering in powerhouses). Reversed, a pump-turbine works as a Francis turbine in pumped-storage plants.

A draft tube is a gradually expanding pipe or passage that connects the runner exit of a reaction turbine to the tailrace, with its outlet always submerged below the tailwater level.

Functions

1. Recovers kinetic energy: the increasing area reduces the high exit velocity, converting kinetic energy into pressure energy instead of wasting it.
2. Creates a negative head (suction) at the runner exit, so the turbine can be installed above the tailwater level without losing head, and the full head $H$ is used.
3. Allows turbine to be set above tailwater for easy inspection and maintenance.
4. Increases turbine output and efficiency (net head on the runner increases).
5. Discharges water safely to the tailrace with low exit velocity.

Pressure at inlet of the draft tube is below atmospheric

     runner
   ===[  ]===   (1) inlet, p1, V1
       \  /        |
        \/         | Hs (above TWL)
 ~~~~~~~~|~~~~~~~~~|~~~~ TWL
        |  |       | y
        |__| (2) outlet, V2

Let $H_s$ = height of draft tube inlet above the tailwater, $y$ = depth of outlet below tailwater, $V_1, V_2$ = velocities at inlet and outlet, $h_f$ = loss in the draft tube, $p_a$ = atmospheric pressure. Take the datum at section 2. Bernoulli between 1 and 2:

$$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} + (H_s + y) = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + 0 + h_f$$

At the outlet, $\dfrac{p_2}{\rho g} = \dfrac{p_a}{\rho g} + y$. Substituting:

$$\begin{aligned} \frac{p_1}{\rho g} + \frac{V_1^2}{2g} + H_s + y &= \frac{p_a}{\rho g} + y + \frac{V_2^2}{2g} + h_f \\ \frac{p_1}{\rho g} &= \frac{p_a}{\rho g} - H_s - \left(\frac{V_1^2 - V_2^2}{2g} - h_f\right) \end{aligned}$$

Since $H_s > 0$ and $V_1 > V_2$ with $\dfrac{V_1^2 - V_2^2}{2g} > h_f$ (the draft tube recovers more energy than it loses), the bracketed term is positive. Hence

$$\frac{p_1}{\rho g} < \frac{p_a}{\rho g}$$

i.e. the pressure at the draft tube inlet (runner exit) is less than atmospheric (a vacuum). This suction is the gain in head given by the draft tube. It must not fall below the vapour pressure, otherwise cavitation occurs; this limits $H_s$ (Thoma's criterion).

Selection of a hydraulic turbine means choosing the type (Pelton, Turgo, Crossflow, Francis, Kaplan/propeller, bulb), number of units, speed and setting that give the highest efficiency and lowest cost for the site.

1. Net head (main criterion)

2. Specific speed

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

($N$ rpm, $P$ kW, $H$ m) relates head, power and speed and is the scientific basis of selection:

A higher $N_s$ gives a smaller, cheaper, faster machine and generator, so the highest $N_s$ allowed by cavitation is chosen.

3. Discharge and its variation (part-load efficiency)

• Pelton and Kaplan have flat efficiency curves and work well at part load (wide flow variation, e.g. RoR plants in dry season).
• Francis and propeller turbines have peaked curves; efficiency falls sharply below 50–60% flow. Several units are used when flow varies widely.

4. Operating speed

Speed should be a synchronous speed $N = \dfrac{120f}{P}$ (50 Hz); higher speed reduces generator cost.

5. Cavitation and turbine setting

Reaction turbines must satisfy Thoma's criterion $\sigma = \dfrac{H_a - H_v - H_s}{H} \ge \sigma_c$. High-$N_s$ turbines need deep setting below tailwater, increasing excavation cost.

6. Sediment and abrasion

Nepalese rivers carry hard quartz silt; Pelton turbines are easier to repair (buckets/needles replaceable) than Francis runners, so this influences selection at medium-high heads.

7. Tailwater level variation

Impulse turbines lose the head between runner and tailwater, so they suit high heads; reaction turbines with draft tubes use the full head and suit low and medium heads with varying tailwater.

8. Other factors

Number of units, overall cost (turbine, generator, powerhouse), runaway speed, ease of maintenance, transport limits in hilly areas, local manufacturing and experience.

An impulse turbine uses only the kinetic energy of a free jet at atmospheric pressure, while a reaction turbine uses both pressure and kinetic energy, with water flowing through the runner under pressure.

Nepalese examples:

• Impulse: Pelton turbines at Khimti (about 680 m head), Upper Tamakoshi (about 820 m head) and Bhotekoshi.
• Reaction: Francis turbines at Kaligandaki A, Middle Marsyangdi and Kulekhani; Kaplan turbines at Trishuli and low-head micro plants.

A draft tube is a gradually diverging tube connecting the runner exit of a Francis (reaction) turbine to the tailrace, with its outlet submerged below tailwater. It is used in a Francis turbine for the following reasons:

1. Recovery of exit kinetic energy: in a Francis turbine water leaves the runner with a considerable velocity (about 5–10 m/s). Without a draft tube this kinetic energy ($V^2/2g$) would be wasted in the tailrace. The diverging tube reduces the velocity and converts most of this energy into pressure energy, increasing the useful head.

2. Creating suction (negative head) below the runner: the column of water in the draft tube produces a pressure below atmospheric at the runner exit:

$$\frac{p_1}{\rho g} = \frac{p_a}{\rho g} - H_s - \left(\frac{V_1^2 - V_2^2}{2g} - h_f\right)$$

So the head between the runner and the tailwater ($H_s$) is not lost; the full head is utilised.

3. Setting above tailwater: the turbine can be placed above the tailwater level, making the powerhouse dry, easy to inspect and maintain, while still using the full available head.

4. Higher efficiency and output: overall efficiency improves by several per cent, which matters greatly for large Francis units.

5. Safe discharge of water to the tailrace at low velocity.

   spiral casing + runner
       ===[  ]===
          \  /   <- inlet: high V, p < p_atm
           ||    Hs
 ~~~~~~~~~ || ~~~~~~~~ TWL
          /  \   diverging tube
         /____\  <- outlet: low V

The height $H_s$ is limited by cavitation (Thoma's criterion), since the inlet pressure must stay above vapour pressure.

Function of a draft tube

A draft tube is a gradually expanding tube between the runner exit of a reaction turbine and the tailrace, with its outlet submerged. Its functions:

1. Converts the high exit kinetic energy into pressure energy (energy recovery), raising efficiency.
2. Produces a negative head (suction) at the runner exit, so the turbine can be set above tailwater without loss of head.
3. Permits the turbine to be installed above tailwater for easy inspection and maintenance.
4. Discharges water to the tailrace at low velocity.

Selection criteria of hydraulic turbines

1. Net head – the primary guide:

2. Specific speed $N_s = \dfrac{N\sqrt{P}}{H^{5/4}}$ (SI: rpm, kW, m):

The highest permissible $N_s$ is chosen because it gives a smaller and cheaper turbine and generator.

3. Discharge and part-load operation: for widely varying flow (RoR plants), Pelton or Kaplan (flat efficiency curves) or several Francis units are chosen; Francis and propeller lose efficiency below about 50–60% load.

4. Speed: must be a synchronous speed $N = 120f/P$ for direct coupling; a high speed reduces generator size.

5. Cavitation and setting: Thoma's criterion $\sigma = \dfrac{H_a - H_v - H_s}{H} \ge \sigma_c$ fixes the suction head. High-$N_s$ machines need deep setting and more excavation.

6. Sediment and abrasion: in silt-laden Nepalese rivers, turbines whose parts are easy to repair or replace (Pelton buckets, needles) are preferred at medium–high heads; coatings are used for Francis runners.

7. Tailwater variation: reaction turbines with draft tubes use the full head even with varying tailwater; impulse turbines lose the setting height.

8. Number of units: depends on load pattern, flow variation, transport size limits, and reliability (at least two units).

9. Cost and maintenance: combined cost of turbine, generator, powerhouse and excavation; availability of spare parts, local expertise and manufacturer.

Objectives of a draft tube for a reaction turbine

A draft tube is a diverging pipe from the runner exit of a reaction turbine to the tailrace, with its outlet submerged. It is provided:

1. To recover kinetic energy at the runner exit by reducing velocity, converting it into pressure energy.
2. To create a suction head (pressure below atmospheric) at the runner exit, so that the head between the runner and tailwater is not lost.
3. To allow the turbine to be set above tailwater for easy access, inspection and repair without loss of head.
4. To increase the net working head, efficiency and output of the turbine.
5. To discharge water to the tailrace with low velocity, preventing scour.

Types: conical (straight divergent), elbow with varying section, simple elbow, Moody spreading tube. Its depth is limited by cavitation (Thoma's criterion).

Working principle of governors

A governor automatically keeps the turbine speed constant by matching the water supply to the load. A conventional oil-pressure (hydraulic) governor has:

• Speed sensor: centrifugal (fly-ball) governor or electronic speed transducer driven from the turbine shaft.
• Control (relay/distributor) valve: directs pressure oil to either side of a servomotor piston.
• Servomotor: a cylinder and piston that moves the guide vanes (Francis/Kaplan) or the spear and deflector (Pelton).
• Oil sump, pump and pressure accumulator: supply oil under pressure.
• Feedback (restoring) mechanism / dashpot: prevents over-correction and hunting.

 turbine shaft -> fly-ball -> lever -> control valve
                                        |   oil
                                        v
  guide vanes / spear <---- servomotor piston
          ^                         |
          |____ feedback linkage ___|

Sequence (load decreases):

1. The turbine speed rises; fly-balls move outward and the sleeve rises.
2. The lever moves the control valve so that pressure oil enters one side of the servomotor.
3. The servomotor piston moves and closes the guide vanes (or pushes the spear forward), reducing discharge.
4. The speed returns to normal; the feedback linkage brings the control valve back to the neutral position.

When the load increases, the reverse happens and the gates open. Modern plants use electro-hydraulic or digital (PID) governors, which sense frequency electronically but still use oil servomotors to move the gates. For Pelton turbines a jet deflector acts first on sudden load rejection to avoid water hammer, while the spear closes slowly.

Given: inlet diameter $D_1 = 1.5$ m, outlet diameter $D_2 = 2.1$ m, outlet velocity $V_2 = 3$ m/s, total length 7.5 m, length immersed in tailwater $y = 1.5$ m, friction loss $h_f = 0.2\,V_2^2/2g$. Take atmospheric pressure head $= 10.3$ m of water.

  (1) inlet  D1 = 1.5 m
     \    /     ^
      \  /      | 6.0 m above TWL
 ~~~~~\  /~~~~~~|~~~~~ TWL
       \/       | 1.5 m below TWL
  (2) outlet D2 = 2.1 m, V2 = 3 m/s

Height of inlet above tailwater $H_s = 7.5 - 1.5 = 6.0$ m.

Velocity at inlet (continuity)

$$\begin{aligned} V_1 &= V_2\left(\frac{D_2}{D_1}\right)^2 = 3 \times \left(\frac{2.1}{1.5}\right)^2 = 5.88\ \text{m/s} \end{aligned}$$

Velocity heads and loss

$$\begin{aligned} \frac{V_1^2}{2g} &= \frac{5.88^2}{19.62} = 1.762\ \text{m} \\ \frac{V_2^2}{2g} &= \frac{3^2}{19.62} = 0.459\ \text{m} \\ h_f &= 0.2 \times 0.459 = 0.092\ \text{m} \end{aligned}$$

Pressure head at inlet

Bernoulli between 1 and 2 (datum at outlet, gauge pressures; $p_2/\rho g = 1.5$ m):

$$\begin{aligned} \frac{p_1}{\rho g} + \frac{V_1^2}{2g} + 7.5 &= \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + h_f \\ \frac{p_1}{\rho g} + 1.762 + 7.5 &= 1.5 + 0.459 + 0.092 \\ \frac{p_1}{\rho g} &= -7.21\ \text{m (gauge)} \end{aligned}$$

Absolute pressure head $= 10.3 - 7.21 = 3.09$ m of water (abs).

Efficiency of the draft tube

$$\begin{aligned} \eta_d &= \frac{\dfrac{V_1^2 - V_2^2}{2g} - h_f}{\dfrac{V_1^2}{2g}} = \frac{1.762 - 0.459 - 0.092}{1.762} \\ &= \frac{1.211}{1.762} = 0.688 \end{aligned}$$

Answer: pressure head at inlet $= -7.21$ m of water (vacuum), i.e. 3.09 m of water absolute; efficiency of draft tube $= 68.8\%$.

Performance characteristic curves show how a turbine's discharge, power and efficiency vary with speed, load or gate opening. Impulse (Pelton) and reaction (Francis, Kaplan) turbines show clearly different curves.

Main characteristics (constant head)

 Efficiency vs % of full load
 eta
  ^   Kaplan  ________
  |   Pelton /--------\_
  |       / /  Francis \
  |      / /  .---.     \
  |     / /  /     \  propeller
  |    / /  /       \
  +-------------------------> % load
     25   50   75  100

Conclusions

• Pelton turbines suit plants with widely varying load and flow at high heads because of their flat efficiency curve.
• Francis turbines give the highest peak efficiency (about 93–95%) but should run near full load; several units are used when flow varies.
• Kaplan turbines, with adjustable blades, keep high efficiency over a wide load range at low heads.

The Pelton turbine is a tangential-flow impulse turbine for high heads, while the Francis turbine is a mixed (radial-inward to axial) flow reaction turbine for medium heads. Their performance curves at constant head differ as follows.

Characteristic curves (constant head)

 (a) eta vs % load          (b) Q vs N (gate const.)
 eta                         Q
  ^   ______ Pelton          ^ ------------ Pelton
  |  /      \_               |\
  | /  .--.   Francis        | \___ Francis
  |/  /    \                 |     \___
  +-----------> % load       +-----------> N
   0  50  100                 (decreases)

Comparison

Interpretation

• Because of its flat efficiency curve, a Pelton turbine is preferred where the flow varies widely (e.g. RoR plants in the dry season).
• The Francis turbine has a higher peak efficiency but needs to run near rated load; for large flow variation, several Francis units are installed and switched on/off.

Purpose of a governor

A governor automatically regulates the flow of water to the turbine so that the speed stays constant (synchronous speed, giving 50 Hz) whatever the load on the generator. It:

• keeps frequency and voltage constant when load changes;
• prevents runaway on load rejection;
• shares load among units in parallel;
• uses water only in proportion to the power demanded.

Governing of a Francis turbine

In a Francis turbine the discharge is controlled by the movable guide vanes (wicket gates) arranged around the runner. All guide vanes are linked to a regulating ring, which is rotated by an oil-pressure servomotor controlled by the governor.

 fly-ball/speed sensor
      |
   control (relay) valve <-- oil pump + accumulator
      |  pressure oil
   servomotor piston
      |
   regulating ring --> guide vanes (open/close)
      |
   feedback linkage -> back to control valve

Working (load decreases):

1. Turbine speed rises above normal; fly-balls move out (or the electronic sensor detects rising frequency).
2. The control valve opens a port, sending pressure oil to one side of the servomotor piston.
3. The piston moves and rotates the regulating ring, closing the guide vanes. The flow area and discharge decrease, and the power input falls to match the load.
4. Speed returns to normal; feedback brings the control valve back to neutral to prevent hunting.

When the load increases, the guide vanes open in the same way.

Relief (pressure regulating) valve: a Francis turbine has no jet deflector, so rapid closure of guide vanes on sudden load rejection would cause high water hammer in the penstock. A relief valve (synchronous bypass) connected to the spiral casing opens at the same time as the guide vanes close, diverting water to the tailrace. It then closes slowly, so the penstock flow is reduced gradually. On long penstocks a surge tank also helps.

Modern Francis units use electro-hydraulic PID governors, where an electronic controller senses frequency and drives the same oil servomotor.

Cavitation is the formation of vapour bubbles in flowing water at points where the local absolute pressure falls to the vapour pressure, and their sudden collapse when they are carried to regions of higher pressure. In reaction turbines it occurs mainly at the runner exit and draft tube inlet, where the pressure is lowest.

Effects of cavitation

1. Pitting and erosion of runner blades, guide vanes and draft tube liner (metal surface becomes rough and spongy); repeated repair is needed.
2. Drop in efficiency and output as bubbles disturb the flow and reduce effective flow area.
3. Noise and vibration of the machine, leading to fatigue of parts.
4. Sudden fall in discharge and power at severe cavitation; in Nepal, cavitation combined with silt abrasion damages runners very quickly.

Avoiding cavitation in reaction turbines

1. Proper setting (suction head) by Thoma's criterion:

$$\sigma = \frac{H_a - H_v - H_s}{H} \ge \sigma_c \;\Rightarrow\; H_s \le H_a - H_v - \sigma_c H$$

where $H_a$ = atmospheric pressure head, $H_v$ = vapour pressure head, $H_s$ = height of runner above tailwater, $H$ = net head, $\sigma_c$ = critical Thoma coefficient. Setting the runner lower (even below tailwater) raises the exit pressure. 2. Selecting a suitable specific speed: high-$N_s$ turbines have higher $\sigma_c$, so they need lower setting. 3. Cavitation-resistant materials: stainless steel (13Cr-4Ni), welded overlays, or coatings on runner blades. 4. Smooth, well-designed blade profiles without sharp edges; polished surfaces. 5. Avoid prolonged part-load or overload operation, which causes vortices in the draft tube. 6. Admit air into the draft tube (air admission valves) to cushion the collapse of bubbles at part load.

Given: $Q = 380$ m³/s, $H = 30$ m, $N = 150$ rpm, overall efficiency $\eta_o = 0.90$, specific speed of Kaplan unit $N_s = 620$ (SI: rpm, kW, m).

Formulas:

$$P = \eta_o\,\rho g Q H, \qquad N_s = \frac{N\sqrt{P}}{H^{5/4}}$$

1. Total power available

$$\begin{aligned} P_{total} &= 0.90 \times 9.81 \times 380 \times 30 \\ &= 100{,}650.6\ \text{kW} \end{aligned}$$

2. Power of one unit from specific speed

$$\begin{aligned} H^{5/4} &= 30^{1.25} = 70.21 \\ \sqrt{P} &= \frac{N_s H^{5/4}}{N} = \frac{620 \times 70.21}{150} = 290.20 \\ P_{unit} &= 290.20^2 = 84{,}218\ \text{kW} \end{aligned}$$

3. Number of units

$$n = \frac{P_{total}}{P_{unit}} = \frac{100{,}650.6}{84{,}218} = 1.195$$

Adopt 2 units (the next whole number, so that the specific speed of each unit does not exceed 620).

4. Output of each unit

$$P_{each} = \frac{100{,}650.6}{2} = 50{,}325.3\ \text{kW} \approx 50.3\ \text{MW}$$

Check: actual specific speed $= \dfrac{150\sqrt{50{,}325.3}}{70.21} = 479$, within the Kaplan range.

Answer: number of Kaplan units $= 2$; output of each unit $\approx 50{,}325$ kW (50.3 MW).

Given: $Q = 340$ m³/s, $H = 27.5$ m, $N = 166.7$ rpm, overall efficiency $\eta_o = 0.88$, specific speed $N_s = 560$ (SI: rpm, kW, m).

Formulas:

$$P = \eta_o\,\rho g Q H, \qquad N_s = \frac{N\sqrt{P}}{H^{5/4}}$$

1. Total power developed

$$\begin{aligned} P_{total} &= 0.88 \times 9.81 \times 340 \times 27.5 \\ &= 80{,}716.7\ \text{kW} \end{aligned}$$

2. Power per unit from specific speed

$$\begin{aligned} H^{5/4} &= 27.5^{1.25} = 62.97 \\ \sqrt{P} &= \frac{N_s H^{5/4}}{N} = \frac{560 \times 62.97}{166.7} = 211.55 \\ P_{unit} &= 211.55^2 = 44{,}754\ \text{kW} \end{aligned}$$

3. Number of units

$$n = \frac{80{,}716.7}{44{,}754} = 1.80$$

Adopt 2 units.

4. Output of each unit and check

$$P_{each} = \frac{80{,}716.7}{2} = 40{,}358\ \text{kW} \approx 40.4\ \text{MW}$$

Actual specific speed $= \dfrac{166.7\sqrt{40{,}358}}{62.97} = 532 < 560$ — acceptable for a Kaplan turbine.

Answer: number of units required $= 2$ Kaplan turbines, each developing about 40,358 kW (40.4 MW).

The number of units is found by comparing the total power available with the power one turbine of the given specific speed can deliver at the given speed.

Data: $Q = 150$ m³/s, $H = 46$ m, $N = 165$ rpm, $\eta_o = 0.865$, $N_s = 234$ (metric, $P$ in kW).

Step 1: Total power available

$$\begin{aligned} P &= \eta_o \, \rho g Q H \\ &= 0.865 \times 9.81 \times 150 \times 46 \ \text{kW} \\ &= 58551.0\ \text{kW} \approx 58.55\ \text{MW} \end{aligned}$$

Step 2: Power of one unit from the specific speed

Specific speed of a turbine is

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

so the power of one unit is

$$\begin{aligned} \sqrt{P_1} &= \frac{N_s H^{5/4}}{N} = \frac{234 \times 46^{1.25}}{165} = \frac{234 \times 119.80}{165} = 169.89 \\ P_1 &= 169.89^2 = 28864\ \text{kW} \end{aligned}$$

Step 3: Number of units

$$n = \frac{P}{P_1} = \frac{58551}{28864} = 2.03 \approx 2$$

Take 2 units (a whole number; 2.03 is practically 2).

Step 4: Output of each unit

$$P_{unit} = \frac{58551}{2} = 29275\ \text{kW} \approx 29.28\ \text{MW}$$

Check: actual specific speed $= \dfrac{165\sqrt{29275}}{46^{1.25}} = 235.7$, very close to the selected 234, so the choice is correct. Discharge per unit $= 150/2 = 75$ m³/s.

Answer: 2 propeller units, each giving about 29275 kW (29.28 MW).

A Francis turbine is designed by fixing its power, a synchronous speed, the specific speed and then the main runner dimensions and the safe setting (suction head) against cavitation. The empirical relations used are the standard ones of de Siervo and de Leva given in the ESHA small hydro guide.

Data: $Q = 55$ m³/s, $H = 25$ m, $\eta = 0.85$, $H_{atm} = 10$ m, $H_v = 0.2$ m of water, frequency $f = 50$ Hz.

1. Power output

$$P = \eta \rho g Q H = 0.85 \times 9.81 \times 55 \times 25 = 11465.4\ \text{kW}$$

2. Trial specific speed and speed

Empirical relation for Francis turbines (metric, $P$ in kW):

$$\begin{aligned} N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{25^{0.625}} = \frac{3470}{7.477} = 464.1 \\ N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{464.1 \times 55.90}{107.08} = 242.3\ \text{rpm} \end{aligned}$$

3. Synchronous speed and actual specific speed

The generator must run at $N = 120f/p$. The nearest synchronous speed is with $p = 24$ poles:

$$\begin{aligned} N &= \frac{120 \times 50}{24} = 250\ \text{rpm} \\ N_s &= \frac{250\sqrt{11465.4}}{25^{1.25}} = 478.9 \end{aligned}$$

This is at the upper end of the Francis range (about 50 to 450, fast runners up to about 500). It is a fast (high specific speed) Francis runner; at such a low head a Kaplan turbine could also be considered, but the design is done for a Francis turbine as asked.

4. Runner outlet (throat) diameter $D_3$

Peripheral velocity coefficient:

$$\begin{aligned} K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 478.9 = 1.507 \\ D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 1.507 \times \sqrt{25}}{250} = 2.547\ \text{m} \end{aligned}$$

5. Runner inlet diameter $D_1$ and inlet height $H_1$

$$\begin{aligned} D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 0.597 \times 2.547 = 1.521\ \text{m} \\ H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.2137 \times 2.547 = 0.544\ \text{m} \end{aligned}$$

$H_1$ is the height of the runner inlet, which is also the height of the guide vanes.

6. Cavitation check and setting of the turbine

Critical Thoma cavitation coefficient for Francis turbines:

$$\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 478.9^{1.41} = 0.4534$$

Maximum suction head (height of runner outlet above tailwater):

$$\begin{aligned} H_s &= H_{atm} - H_v - \sigma_c H \\ &= 10 - 0.2 - 0.4534 \times 25 \\ &= -1.53\ \text{m} \end{aligned}$$

The negative sign means the runner outlet must be placed 1.53 m below the minimum tailwater level (submerged setting) to avoid cavitation.

Summary of design

A spiral casing, 16 to 24 guide vanes and an elbow draft tube starting at diameter about $D_3$ complete the unit.

A Francis turbine is designed by fixing its power, a synchronous speed, the specific speed and then the main runner dimensions and the safe setting (suction head) against cavitation. The empirical relations used are the standard ones of de Siervo and de Leva given in the ESHA small hydro guide.

Data: $Q = 45$ m³/s, $H = 150$ m, $\eta = 0.85$, $H_{atm} = 10.3$ m, $H_v = 0.2$ m of water, frequency $f = 50$ Hz.

1. Power output

$$P = \eta \rho g Q H = 0.85 \times 9.81 \times 45 \times 150 = 56284.9\ \text{kW}$$

2. Trial specific speed and speed

Empirical relation for Francis turbines (metric, $P$ in kW):

$$\begin{aligned} N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{150^{0.625}} = \frac{3470}{22.912} = 151.5 \\ N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{151.5 \times 524.95}{237.24} = 335.1\ \text{rpm} \end{aligned}$$

3. Synchronous speed and actual specific speed

The generator must run at $N = 120f/p$. The nearest synchronous speed is with $p = 18$ poles:

$$\begin{aligned} N &= \frac{120 \times 50}{18} = 333.33\ \text{rpm} \\ N_s &= \frac{333.33\sqrt{56284.9}}{150^{1.25}} = 150.6 \end{aligned}$$

This lies in the Francis range (about 50 to 450).

4. Runner outlet (throat) diameter $D_3$

Peripheral velocity coefficient:

$$\begin{aligned} K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 150.6 = 0.687 \\ D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 0.687 \times \sqrt{150}}{333.33} = 2.132\ \text{m} \end{aligned}$$

5. Runner inlet diameter $D_1$ and inlet height $H_1$

$$\begin{aligned} D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 1.027 \times 2.132 = 2.190\ \text{m} \\ H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.1317 \times 2.132 = 0.281\ \text{m} \end{aligned}$$

$H_1$ is the height of the runner inlet, which is also the height of the guide vanes.

6. Cavitation check and setting of the turbine

Critical Thoma cavitation coefficient for Francis turbines:

$$\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 150.6^{1.41} = 0.0888$$

Maximum suction head (height of runner outlet above tailwater):

$$\begin{aligned} H_s &= H_{atm} - H_v - \sigma_c H \\ &= 10.3 - 0.2 - 0.0888 \times 150 \\ &= -3.22\ \text{m} \end{aligned}$$

The negative sign means the runner outlet must be placed 3.22 m below the minimum tailwater level (submerged setting) to avoid cavitation.

Summary of design

A spiral casing, 16 to 24 guide vanes and an elbow draft tube starting at diameter about $D_3$ complete the unit.

A Francis turbine is designed by fixing its power, a synchronous speed, the specific speed and then the main runner dimensions and the safe setting (suction head) against cavitation. The empirical relations used are the standard ones of de Siervo and de Leva given in the ESHA small hydro guide. Atmospheric and vapour pressure heads are not given, so standard values are assumed.

Data: $Q = 140$ m³/s, $H = 120$ m, $\eta = 0.94$, $H_{atm} = 10.3$ m, $H_v = 0.2$ m of water (assumed: sea-level atmosphere and water at about 20°C), frequency $f = 50$ Hz.

1. Power output

$$P = \eta \rho g Q H = 0.94 \times 9.81 \times 140 \times 120 = 154919.5\ \text{kW}$$

2. Trial specific speed and speed

Empirical relation for Francis turbines (metric, $P$ in kW):

$$\begin{aligned} N_s &= \frac{3470}{H^{0.625}} = \frac{3470}{120^{0.625}} = \frac{3470}{19.929} = 174.1 \\ N &= \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{174.1 \times 397.17}{393.60} = 175.7\ \text{rpm} \end{aligned}$$

3. Synchronous speed and actual specific speed

The generator must run at $N = 120f/p$. The nearest synchronous speed is with $p = 34$ poles:

$$\begin{aligned} N &= \frac{120 \times 50}{34} = 176.47\ \text{rpm} \\ N_s &= \frac{176.47\sqrt{154919.5}}{120^{1.25}} = 174.9 \end{aligned}$$

This lies in the Francis range (about 50 to 450). One unit of about 155 MW is designed here; in practice the flow may be shared between two or more units of the same design.

4. Runner outlet (throat) diameter $D_3$

Peripheral velocity coefficient:

$$\begin{aligned} K_u &= 0.31 + 0.0025 N_s = 0.31 + 0.0025 \times 174.9 = 0.747 \\ D_3 &= \frac{84.5 K_u \sqrt{H}}{N} = \frac{84.5 \times 0.747 \times \sqrt{120}}{176.47} = 3.919\ \text{m} \end{aligned}$$

5. Runner inlet diameter $D_1$ and inlet height $H_1$

$$\begin{aligned} D_1 &= \left(0.4 + \frac{94.5}{N_s}\right) D_3 = 0.940 \times 3.919 = 3.686\ \text{m} \\ H_1 &= (0.094 + 0.00025 N_s) D_3 = 0.1377 \times 3.919 = 0.540\ \text{m} \end{aligned}$$

$H_1$ is the height of the runner inlet, which is also the height of the guide vanes.

6. Cavitation check and setting of the turbine

Critical Thoma cavitation coefficient for Francis turbines:

$$\sigma_c = 7.54 \times 10^{-5} N_s^{1.41} = 7.54 \times 10^{-5} \times 174.9^{1.41} = 0.1096$$

Maximum suction head (height of runner outlet above tailwater):

$$\begin{aligned} H_s &= H_{atm} - H_v - \sigma_c H \\ &= 10.3 - 0.2 - 0.1096 \times 120 \\ &= -3.05\ \text{m} \end{aligned}$$

The negative sign means the runner outlet must be placed 3.05 m below the minimum tailwater level (submerged setting) to avoid cavitation.

Summary of design

A spiral casing, 16 to 24 guide vanes and an elbow draft tube starting at diameter about $D_3$ complete the unit.

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets.

Data: $Q = 3.75$ m³/s, $H = 380$ m, $\eta_o = 0.86$, $f = 50$ Hz. Given: coefficient of velocity of the jet $C_v = 0.97$ and speed ratio $\phi = 0.46$.

1. Power developed

$$P = \eta_o \rho g Q H = 0.86 \times 9.81 \times 3.75 \times 380 = 12022.2\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 380} = 86.35\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.97 \times 86.35 = 83.76\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 86.35 = 39.72\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 3.75}{\pi \times 83.76}} = 0.2388\ \text{m} \approx 239\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 2.865$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 39.72}{\pi \times 2.865} = 264.8\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/22 = 272.73$ rpm (22 poles).

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 39.72}{\pi \times 272.73} = 2.781\ \text{m} \\ m &= \frac{D}{d} = \frac{2.781}{0.2388} = 11.65 \end{aligned}$$

$m$ lies within 10 to 14, so a single-jet wheel is satisfactory.

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{272.73 \times 109.65}{1677.8} = 17.8$$

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

7. Number of buckets (Tygun's formula)

$$z = 15 + \frac{D}{2d} = 15 + \frac{11.65}{2} = 20.82 \approx 21$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 1.194$ m
• Bucket depth $T \approx 1.2d = 0.287$ m
• Bucket radial length $L \approx 2.5d = 0.597$ m

Summary of design

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets.

Data: $Q = 2$ m³/s, $H = 500$ m, $\eta_o = 0.85$, $f = 50$ Hz. Assumed from the standard ranges: $C_v = 0.98$ (0.97 to 0.99) and speed ratio $\phi = 0.46$ (0.43 to 0.47).

1. Power developed

$$P = \eta_o \rho g Q H = 0.85 \times 9.81 \times 2 \times 500 = 8338.5\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 500} = 99.05\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 99.05 = 97.06\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 99.05 = 45.56\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 2}{\pi \times 97.06}} = 0.1620\ \text{m} \approx 162\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 1.944$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 45.56}{\pi \times 1.944} = 447.7\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/14 = 428.57$ rpm (14 poles).

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 45.56}{\pi \times 428.57} = 2.030\ \text{m} \\ m &= \frac{D}{d} = \frac{2.030}{0.1620} = 12.54 \end{aligned}$$

$m$ lies within 10 to 14, so a single-jet wheel is satisfactory.

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{428.57 \times 91.32}{2364.4} = 16.6$$

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

7. Number of buckets (Tygun's formula)

$$z = 15 + \frac{D}{2d} = 15 + \frac{12.54}{2} = 21.27 \approx 22$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 0.810$ m
• Bucket depth $T \approx 1.2d = 0.194$ m
• Bucket radial length $L \approx 2.5d = 0.405$ m

Summary of design

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets. The given velocity coefficient 0.98 is used for the jet.

Data: $Q = 5$ m³/s, $H = 312.5$ m, $\eta_o = 0.85$, $f = 50$ Hz. Assumed from the standard ranges: $C_v = 0.98$ (0.97 to 0.99) and speed ratio $\phi = 0.46$ (0.43 to 0.47).

1. Power developed

$$P = \eta_o \rho g Q H = 0.85 \times 9.81 \times 5 \times 312.5 = 13028.9\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 312.5} = 78.30\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 78.30 = 76.74\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 78.30 = 36.02\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 5}{\pi \times 76.74}} = 0.2880\ \text{m} \approx 288\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 3.456$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 36.02}{\pi \times 3.456} = 199.0\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/30 = 200$ rpm (30 poles).

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 36.02}{\pi \times 200} = 3.440\ \text{m} \\ m &= \frac{D}{d} = \frac{3.440}{0.2880} = 11.94 \end{aligned}$$

$m$ lies within 10 to 14, so a single-jet wheel is satisfactory.

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{200 \times 114.14}{1313.9} = 17.4$$

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

7. Number of buckets (Tygun's formula)

$$z = 15 + \frac{D}{2d} = 15 + \frac{11.94}{2} = 20.97 \approx 21$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 1.440$ m
• Bucket depth $T \approx 1.2d = 0.346$ m
• Bucket radial length $L \approx 2.5d = 0.720$ m

Summary of design

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets.

Data: $Q = 5$ m³/s, $H = 310$ m, $\eta_o = 0.86$, $f = 50$ Hz. Assumed from the standard ranges: $C_v = 0.98$ (0.97 to 0.99) and speed ratio $\phi = 0.46$ (0.43 to 0.47).

1. Power developed

$$P = \eta_o \rho g Q H = 0.86 \times 9.81 \times 5 \times 310 = 13076.7\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 310} = 77.99\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 77.99 = 76.43\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 77.99 = 35.87\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 5}{\pi \times 76.43}} = 0.2886\ \text{m} \approx 289\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 3.463$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 35.87}{\pi \times 3.463} = 197.8\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/30 = 200$ rpm (30 poles).

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 35.87}{\pi \times 200} = 3.426\ \text{m} \\ m &= \frac{D}{d} = \frac{3.426}{0.2886} = 11.87 \end{aligned}$$

$m$ lies within 10 to 14, so a single-jet wheel is satisfactory.

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{200 \times 114.35}{1300.8} = 17.6$$

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

7. Number of buckets (Tygun's formula)

$$z = 15 + \frac{D}{2d} = 15 + \frac{11.87}{2} = 20.93 \approx 21$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 1.443$ m
• Bucket depth $T \approx 1.2d = 0.346$ m
• Bucket radial length $L \approx 2.5d = 0.722$ m

Summary of design

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets. The specific speed given in the data is checked first.

Data: $Q = 2$ m³/s, $H = 605$ m, $\eta_o = 0.87$, $f = 50$ Hz. Assumed from the standard ranges: $C_v = 0.98$ (0.97 to 0.99) and speed ratio $\phi = 0.46$ (0.43 to 0.47).

1. Power developed

$$P = \eta_o \rho g Q H = 0.87 \times 9.81 \times 2 \times 605 = 10327.0\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 605} = 108.95\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 108.95 = 106.77\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 108.95 = 50.12\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 2}{\pi \times 106.77}} = 0.1544\ \text{m} \approx 154\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 1.853$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 50.12}{\pi \times 1.853} = 516.5\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/12 = 500$ rpm (12 poles).

Use of the given $N_s = 100$: if the speed is taken from the given specific speed,

$$N = \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{100 \times 3000.5}{101.62} = 2953\ \text{rpm} \approx 3000\ \text{rpm}$$

which gives $D = 60u/(\pi N) = 0.319$ m and a jet ratio $m = D/d = 2.07$, far below the minimum of about 10. A single jet Pelton has $N_s$ of only about 8.5 to 30; even 6 jets raise this only by $\sqrt{6}$ (to about 70). Reaching 100 would need about $(100/16.9)^2 \approx 35$ jets, which is impossible. So $N_s = 100$ cannot be met by a Pelton wheel at this head; the design below uses the practical speed of 500 rpm and the standard proportions, and the actual $N_s$ is reported.

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 50.12}{\pi \times 500} = 1.914\ \text{m} \\ m &= \frac{D}{d} = \frac{1.914}{0.1544} = 12.40 \end{aligned}$$

$m$ lies within 10 to 14, so a single-jet wheel is satisfactory.

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{500 \times 101.62}{3000.5} = 16.9$$

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

7. Number of buckets (Tygun's formula)

$$z = 15 + \frac{D}{2d} = 15 + \frac{12.40}{2} = 21.20 \approx 22$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 0.772$ m
• Bucket depth $T \approx 1.2d = 0.185$ m
• Bucket radial length $L \approx 2.5d = 0.386$ m

Summary of design