11.4 Hydraulic Turbines, Centrifugal & Reciprocating Pumps

Hydraulic Turbines, Centrifugal & Reciprocating Pumps

Hydraulic machinery converts hydraulic energy into mechanical shaft work (turbines) or inputs shaft work to boost fluid pressure and elevation (pumps). In Coal India operations, heavy-duty centrifugal slurry and multistage mine drainage pumps dewater underground mine sumps and pit bottoms against high static heads, while reciprocating positive-displacement pumps inject high-pressure hydraulic fluid into powered longwall roof supports.

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1. Euler's Turbomachine Equation & Velocity Triangles

                   Inlet Velocity Triangle (Turbine Runner):
                        V_1 (Absolute Velocity)
                       /|\
                      / | \
         (Relative)  /  |  \ V_f1 (Flow Velocity)
               V_r1 /   |   \
                   /____|____\ (alpha_1 = Guide Vane Angle, beta_1 = Blade Angle)
                  <-- u_1 --->
                  <------- V_w1 -------> (Whirl Velocity)

1.1 Velocity Triangle Nomenclature:

• $V$: Absolute velocity of fluid ($\text{m/s}$)
• $u = \frac{\pi D N}{60}$: Tangential peripheral blade velocity ($\text{m/s}$)
• $V_r = \vec{V} - \vec{u}$: Relative velocity of fluid with respect to blade
• $V_w = V\cos\alpha$: Whirl (tangential) component of absolute velocity (produces angular momentum)
• $V_f = V\sin\alpha$: Flow (axial or radial) component of absolute velocity (governs discharge $Q = A V_f$)
• $\alpha$: Guide vane / nozzle angle with tangential direction
• $\beta$: Runner / impeller blade angle with tangential direction

1.2 Euler's Equation for Energy Transfer:

Applying the Angular Momentum Principle (Torque $T = \dot{m}(r_1 V_{w1} - r_2 V_{w2})$):

1. For Hydraulic Turbines (Work extracted from fluid): $W = \dot{m}(u_1 V_{w1} - u_2 V_{w2}) = \rho Q (u_1 V_{w1} - u_2 V_{w2})$ $H_{\text{Euler}} = \frac{u_1 V_{w1} - u_2 V_{w2}}{g}$
2. For Pumps (Work delivered to fluid): $W_{\text{pump}} = \rho Q (u_2 V_{w2} - u_1 V_{w1})$ $H_{\text{Euler}} = \frac{u_2 V_{w2} - u_1 V_{w1}}{g}$

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2. Pelton Wheel (Impulse Turbine)

The Pelton Wheel is a tangential-flow impulse turbine designed for high head ($H > 250\text{ m}$) and low discharge ($Q$).

      High Pressure Water
      from Penstock ======>> [ Nozzle with Spear ]
                                     |
                                     v High Velocity Jet (V_1 = C_v * sqrt(2gH))
                              (( Bucket ))
                             //          \\
                            ||   Runner   ||
                             \\          //
                              (( Bucket ))

2.1 Operating Principle & Jet Mechanics

• All available pressure head is completely converted into kinetic energy at atmospheric pressure across the nozzle spear valve: $V_1 = C_v \sqrt{2gH}$ Where $C_v \approx 0.98 - 0.99$ is the nozzle velocity coefficient.
• Speed Ratio ($\phi$): $\phi = \frac{u_1}{\sqrt{2gH}} \approx 0.44 - 0.48 \quad (\text{Theoretically } 0.50)$
• Runner Bucket Deflection Angle: Double-hemispherical bucket with central splitter ridge. Jet deflection angle $\theta \approx 165^{\circ} - 170^{\circ}$, leaving bucket at outlet angle $\beta_2 = 180^{\circ} - \theta \approx 10^{\circ} - 15^{\circ}$.

2.2 Theoretical Hydraulic Efficiency ($\eta_h$)

Assuming relative velocity reduction factor $k = V_{r2}/V_{r1} \le 1$ ($k = 1$ for frictionless bucket):

$\eta_h = \frac{\text{Work Done/sec}}{\text{Jet Kinetic Energy/sec}} = \frac{2 u (V_1 - u)(1 + k \cos\beta_2)}{V_1^2}$

• Maximum Hydraulic Efficiency Condition: Differentiating $\eta_h$ with respect to $u$ and setting $\frac{d\eta_h}{du} = 0$: $u = \frac{V_1}{2}$ $\eta_{h,\max} = \frac{1 + k \cos\beta_2}{2}$ For frictionless bucket ($k = 1$) and ideal $180^{\circ}$ reversal ($\beta_2 = 0^{\circ} \implies \cos 0^{\circ} = 1$): $\eta_{h,\max} = \frac{1 + 1}{2} = 100\%$ (In practice, $\beta_2 \approx 15^{\circ}$ prevents the exiting jet from hitting the back of the advancing bucket, giving $\eta_{h,\max} \approx 94% - 96%$).

2.3 Pelton Wheel Geometric Design Ratios

• Jet Ratio ($m$): Ratio of pitch diameter of runner ($D$) to jet diameter ($d$): $m = \frac{D}{d} \approx 11 - 16 \quad (\text{Standard design default } m = 12)$
• Number of Buckets on Runner ($Z$ - Tygun's Empirical Formula): $Z = 15 + \frac{m}{2} = 15 + \frac{D}{2d}$

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3. Francis Reaction Turbine

The Francis Turbine is an inward mixed-flow reaction turbine designed for medium head ($60\text{ m} < H < 250\text{ m}$) and medium discharge.

               Spiral Scroll Casing
               +-----------------------+
               |   Stay Ring           |
               |   Guide Vanes (alpha) |
               |   Runner Blades       |
               |      (Radial Inward)  |
               |           |           |
               |           v           |
               |      Draft Tube       |
               +-----------------------+

3.1 Velocity Triangles & Radial Discharge Condition

For maximum hydraulic efficiency in Francis turbines, water discharges radially from the runner outlet:

$\alpha_2 = 90^{\circ} \implies V_{w2} = 0, \quad V_2 = V_{f2}$

Under radial discharge at exit, Euler's head simplifies to:

$H_e = \frac{u_1 V_{w1}}{g}$

$\eta_h = \frac{u_1 V_{w1}}{gH}$

3.2 Design Ratios:

• Flow Ratio ($\psi$): $\psi = \frac{V_{f1}}{\sqrt{2gH}} \approx 0.15 - 0.30$
• Speed Ratio ($\phi$): $\phi = \frac{u_1}{\sqrt{2gH}} \approx 0.60 - 0.90$

3.3 The Draft Tube

A gradually expanding airtight conduit connecting the runner exit (section 2) to the tailrace (section 3).

Functions of Draft Tube:

1. Allows the turbine to be installed above the tailrace level without losing head (creates suction head $z_2$ at runner exit).
2. Recovers a large fraction of the kinetic energy $\frac{V_2^2}{2g}$ rejected by the runner, converting it into useful pressure head by deceleration.

Draft Tube Efficiency ($\eta_d$):

$\eta_d = \frac{\text{Actual Kinetic Head Converted into Pressure Head}}{\text{Kinetic Head entering Draft Tube}} = \frac{\frac{V_2^2 - V_3^2}{2g} - h_L}{\frac{V_2^2}{2g}}$ To prevent boundary layer separation, standard straight conical draft tubes have a maximum total divergence angle of $\le 8^{\circ}$.

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4. Kaplan Turbine (Axial Flow Reaction Turbine)

The Kaplan Turbine is an axial-flow reaction turbine with adjustable runner blades, designed for low head ($H < 60\text{ m}$) and very high discharge ($Q$).

                 Hub / Boss (d_h)
                 +-------+
                 | ##### | <--- Adjustable Blades (4 to 8 blades)
                 | ##### |
                 +-------+
                 <--- D_o ---> (Outer Tip Diameter)

• Axial Flow: Water flows parallel to the axis of rotation throughout: $V_{f1} = V_{f2} = V_f$.
• Discharge Equation: $Q = \frac{\pi}{4} (D_o^2 - d_h^2) V_f$ Where $D_o$ is outer runner tip diameter, $d_h$ is hub/boss diameter ($d_h/D_o \approx 0.35 - 0.60$).
• Kaplan vs. Propeller Turbine: Propeller turbines have fixed runner blades, resulting in sharp efficiency drop at off-design loads. Kaplan turbines have simultaneously adjustable guide vanes and runner blades, maintaining a broad, flat efficiency curve ($> 90%$) over $30% - 100%$ load range.

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5. Turbine Selection, Specific Speed & Unit Quantities

5.1 Specific Speed of Turbines ($N_{s,\text{turbine}}$)

The speed of a geometrically similar model turbine producing unit shaft power ($1\text{ kW}$) under unit net head ($1\text{ m}$):

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

(Where $N$ is in rpm, $P$ is in kW, $H$ is in meters).

5.2 Selection Guide based on Specific Speed:

5.3 Unit Quantities (Performance Scaling under Varying Head):

• Unit Speed ($N_u$): $N_u = \frac{N}{\sqrt{H}}$
• Unit Discharge ($Q_u$): $Q_u = \frac{Q}{\sqrt{H}}$
• Unit Power ($P_u$): $P_u = \frac{P}{H^{3/2}}$

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6. Centrifugal Pumps

A centrifugal pump operates as a dynamic turbomachine transferring shaft work to fluid via centrifugal action (a reversed reaction turbine).

                   Centrifugal Impeller Vane Curvatures:
       1. Backward-Curved Vane:    2. Radial Vane:     3. Forward-Curved Vane:
          (beta_2 < 90 deg)          (beta_2 = 90 deg)     (beta_2 > 90 deg)
               )                          |                     (
          Stable Drooping H-Q        Flat H-Q              Unstable Rising H-Q
          Non-overloading Power      Industrial Blowers    High Kinetic Loss

6.1 Blade Curvature & Backward Curved Vanes ($\beta_2 < 90^{\circ}$)

From the outlet velocity triangle with radial inlet ($V_{w1} = 0$):

$H_m = \frac{u_2 V_{w2}}{g} = \frac{u_2 (u_2 - V_{f2}\cot\beta_2)}{g} = \frac{u_2^2}{g} - \left(\frac{u_2}{g \pi D_2 b_2}\right) Q \cot\beta_2$

• Backward Curved Vanes ($\beta_2 < 90^{\circ} \implies \cot\beta_2 > 0$): $H_m$ decreases linearly with increasing discharge $Q$ (drooping head-discharge characteristic). This produces self-regulating, non-overloading power demand, highest hydraulic efficiency, and stable parallel operation. All commercial centrifugal water and slurry pumps use backward curved vanes.
• Forward Curved Vanes ($\beta_2 > 90^{\circ} \implies \cot\beta_2 < 0$): Head increases with discharge, causing motor overloading at high flows and hydraulic surging/instability.

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6.2 Manometric Head & Efficiencies

• Manometric Head ($H_m$): Net actual head developed across pump flanges: $H_m = h_s + h_d + h_{fs} + h_{fd} + \frac{V_d^2}{2g} = \left(\frac{P_d}{\rho g} - \frac{P_s}{\rho g}\right) + (z_d - z_s) + \frac{V_d^2 - V_s^2}{2g}$
• Manometric Efficiency ($\eta_{\text{mano}}$): $\eta_{\text{mano}} = \frac{g H_m}{u_2 V_{w2}}$
• Overall Efficiency ($\eta_o$): $\eta_o = \frac{\text{Fluid Water Power}}{\text{Shaft Power Input}} = \frac{\rho g Q H_m}{P_{\text{shaft}}} = \eta_{\text{mano}} \times \eta_{\text{vol}} \times \eta_{\text{mech}}$
• Specific Speed of Pumps ($N_{s,\text{pump}}$): $N_s = \frac{N \sqrt{Q}}{H_m^{3/4}}$ (Where $N$ is rpm, $Q$ is $\text{m}^3/\text{s}$, $H_m$ is meters).

6.3 Minimum Starting Speed of Centrifugal Pump

To initiate delivery flow against manometric head $H_m$, the centrifugal head developed across impeller diameters $D_1$ and $D_2$ must at least balance $H_m$:

$\frac{u_2^2 - u_1^2}{2g} \ge H_m = \eta_{\text{mano}} \frac{u_2 V_{w2}}{g}$

$N_{\min} = \frac{120 \, \eta_{\text{mano}} \, V_{w2} \, D_2}{\pi (D_2^2 - D_1^2)}$

6.4 Multistage Centrifugal Pumps

• Pumps in Series (Multi-Impellers on Single Shaft): High head mine dewatering ($H_{\text{total}} = n \times H_m$, $Q$ constant).
• Pumps in Parallel (Multiple Pumps Discharging to Common Header): High volume dewatering ($Q_{\text{total}} = n \times Q$, $H_m$ constant).

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7. Cavitation & Net Positive Suction Head (NPSH)

7.1 Cavitation Phenomenon

If local absolute static pressure at any point (e.g., turbine blade suction side, pump impeller eye) drops below the saturation vapor pressure of the liquid ($P < P_v$), liquid vaporizes instantly into vapor bubbles. When these bubbles are swept into downstream high-pressure zones, they collapse violently within microseconds, generating localized micro-jets with shock pressures exceeding $1000\text{ MPa}$, causing severe pitting fatigue, vibration, noise, and steep drop in efficiency.

     Low Pressure Zone (P < P_v)           High Pressure Zone (P > P_v)
       O o . (Vapor Cavity Bubble Forms)  ===>  (Bubble Implodes -> 1000 MPa Micro-Jet -> Pitting Pits)

7.2 Net Positive Suction Head (NPSH)

$\text{NPSH} = \left(\frac{P_s}{\rho g} + \frac{V_s^2}{2g}\right) - \frac{P_v}{\rho g} = \frac{P_{\text{atm}}}{\rho g} - \frac{P_v}{\rho g} - h_s - h_{fs} = H_{\text{atm}} - H_v - h_s - h_{fs}$

• Cavitation Prevention Criterion: $\text{NPSH}_{\text{available}} > \text{NPSH}_{\text{required}}$
• Thoma's Cavitation Parameter ($\sigma$): $\sigma = \frac{\text{NPSH}}{H} = \frac{H_{\text{atm}} - H_v - h_s}{H}$ Cavitation occurs when $\sigma \le \sigma_c$ (Critical Thoma Parameter).

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8. Reciprocating Pumps & Air Vessels

Reciprocating pumps are positive-displacement machines suitable for high heads and low flow rates.

8.1 Theoretical Discharge & Slip

• Single-Acting Pump: $Q_{\text{th}} = \frac{A L N}{60}$
• Double-Acting Pump: $Q_{\text{th}} = \frac{(2A - a) L N}{60} \approx \frac{2 A L N}{60}$
• Slip: $\text{Slip} = \frac{Q_{\text{th}} - Q_{\text{act}}}{Q_{\text{th}}} \times 100\%$ $C_d = \frac{Q_{\text{act}}}{Q_{\text{th}}} = 1 - \text{Slip}$
• Negative Slip ($Q_{\text{act}} > Q_{\text{th}} \implies \text{Slip} < 0, C_d > 1$): Occurs when the delivery pipe is short, suction pipe is long, and pump operates at high speed, such that suction liquid inertia forces the delivery valve open before the suction stroke finishes.

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8.2 Kinematic Heads: Acceleration and Friction

1. Acceleration Head ($h_a$): $h_a = \frac{l}{g} \left(\frac{A}{a}\right) \omega^2 r \cos\theta$
   • Maximum at stroke ends: $\theta = 0^{\circ} \implies h_{a,\max} = \frac{l}{g} \left(\frac{A}{a}\right) \omega^2 r$; zero at mid-stroke ($\theta = 90^{\circ}$).
2. Friction Head ($h_f$): $h_f = \frac{4 f l}{2 g d} \left(\frac{A}{a} \omega r \sin\theta\right)^2$
   • Maximum at mid-stroke: $\theta = 90^{\circ}$; zero at stroke ends ($\theta = 0^{\circ}, 180^{\circ}$).

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8.3 Function of Air Vessels & Friction Work Savings

An Air Vessel is a cast-iron chamber containing compressed air at top and liquid at bottom, installed adjacent to the cylinder on suction and delivery pipes.

Primary Functions:

1. Converts pulsating, intermittent cylinder discharge into a continuous, uniform pipe flow.
2. Eliminates acceleration head over the main pipeline length beyond the air vessel.
3. Enables the pump to run at significantly higher speeds without cavitation separation.

Percentage Work Saved against Friction by Installing Air Vessels:

    1. Single-Acting Pump (Without Vessel):    With Air Vessel:
       Sinusoidal Velocity Profile              Constant Uniform Velocity
       Friction Work ~ Integral(v^3 dt)         Friction Work ~ v_avg^2 * L
       ===> Work Saved = 84.8%                  ===> Frictional dissipation slashed!

• For a Single-Acting Reciprocating Pump: $\% \text{ Work Saved} = \left(1 - \frac{\pi^2}{8}\right) \dots = \frac{\frac{2}{3} - \frac{1}{\pi^2}}{\frac{2}{3}} \approx \mathbf{84.8\%}$
• For a Double-Acting Reciprocating Pump: $\% \text{ Work Saved} = \mathbf{39.2\%}$

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9. Worked Engineering Examples

Problem 1: Pelton Wheel Hydraulic Efficiency and Power Calculation

A Pelton wheel develops $P_{\text{shaft}} = 6000\text{ kW}$ under a net head of $H = 300\text{ m}$ running at $N = 500\text{ rpm}$. The bucket speed ratio is $\phi = 0.46$, nozzle velocity coefficient $C_v = 0.98$, bucket deflection angle $\theta = 165^{\circ}$ ($\beta_2 = 15^{\circ}$), and relative velocity is reduced by $12%$ across the bucket due to friction ($k = 0.88$). Mechanical efficiency $\eta_{\text{mech}} = 90%$. Calculate:

1. The jet velocity ($V_1$) and runner pitch diameter ($D$).
2. The theoretical hydraulic efficiency ($\eta_h$).
3. The required water flow rate ($Q$).

Solution:

1. Jet Velocity ($V_1$) and Runner Peripheral Velocity ($u$): $V_1 = C_v \sqrt{2gH} = 0.98 \times \sqrt{2 \times 9.81 \times 300} = 0.98 \times 76.72 = 75.186\text{ m/s}$ $u = \phi \sqrt{2gH} = 0.46 \times 76.72 = 35.29\text{ m/s}$
2. Runner Pitch Diameter ($D$): $u = \frac{\pi D N}{60} \implies D = \frac{60 u}{\pi N} = \frac{60 \times 35.29}{\pi \times 500} = \frac{2117.4}{1570.8} = 1.348\text{ m}$
3. Hydraulic Efficiency ($\eta_h$): $\eta_h = \frac{2 u (V_1 - u)(1 + k \cos\beta_2)}{V_1^2}$ $V_1 - u = 75.186 - 35.29 = 39.896\text{ m/s}$ $1 + k \cos 15^{\circ} = 1 + 0.88 \times 0.9659 = 1 + 0.850 = 1.850$ $\eta_h = \frac{2 \times 35.29 \times 39.896 \times 1.850}{(75.186)^2} = \frac{5209.6}{5652.9} = 0.9216 = 92.16\%$
4. Water Flow Rate ($Q$): $\eta_{\text{overall}} = \eta_h \times \eta_{\text{mech}} = 0.9216 \times 0.90 = 0.8294 = 82.94\%$ $P_{\text{shaft}} = \eta_{\text{overall}} \cdot \rho g Q H$ $6000 \times 10^3 = 0.8294 \times 1000 \times 9.81 \times Q \times 300$ $Q = \frac{6000 \times 10^3}{2,440,924} = 2.458\text{ m}^3/\text{s}$

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Problem 2: Centrifugal Mine Dewatering Pump

A centrifugal pump is required to lift $Q = 0.05\text{ m}^3/\text{s}$ of water against a total manometric head of $H_m = 40\text{ m}$. The impeller diameter is $D_2 = 300\text{ mm}$, width at outlet $b_2 = 25\text{ mm}$, running at $N = 1450\text{ rpm}$. The outlet blade angle is backward-curved at $\beta_2 = 30^{\circ}$. Assuming radial entry at inlet, determine:

1. The whirl velocity at outlet ($V_{w2}$).
2. The manometric efficiency ($\eta_{\text{mano}}$).
3. The shaft input power if mechanical efficiency is $92%$.

Solution:

1. Peripheral Velocity ($u_2$) and Flow Velocity ($V_{f2}$): $u_2 = \frac{\pi D_2 N}{60} = \frac{\pi \times 0.30 \times 1450}{60} = 22.776\text{ m/s}$ $A_2 = \pi D_2 b_2 = \pi \times 0.30 \times 0.025 = 0.02356\text{ m}^2$ $V_{f2} = \frac{Q}{A_2} = \frac{0.05}{0.02356} = 2.122\text{ m/s}$
2. Whirl Velocity ($V_{w2}$): $V_{w2} = u_2 - V_{f2}\cot\beta_2 = 22.776 - 2.122 \times \cot 30^{\circ} = 22.776 - 2.122 \times 1.732 = 22.776 - 3.675 = 19.101\text{ m/s}$
3. Manometric Efficiency ($\eta_{\text{mano}}$): $\eta_{\text{mano}} = \frac{g H_m}{u_2 V_{w2}} = \frac{9.81 \times 40.0}{22.776 \times 19.101} = \frac{392.4}{435.05} = 0.9019 = 90.19\%$
4. Shaft Input Power ($P_{\text{shaft}}$): $P_{\text{fluid}} = \rho g Q H_m = 1000 \times 9.81 \times 0.05 \times 40.0 = 19,620\text{ W} = 19.62\text{ kW}$ $P_{\text{shaft}} = \frac{P_{\text{fluid}}}{\eta_{\text{mano}} \times \eta_{\text{mech}}} = \frac{19.62}{0.9019 \times 0.92} = \frac{19.62}{0.8297} = 23.65\text{ kW}$ The motor must deliver $23.65\text{ kW}$ to the pump shaft.