MEEN 20010 – Mechanics of Fluids I – Laboratory No. 2 – Analysis of a Pelton Wheel Water Turbine PERFORMANCE ANALYSIS OF A PELTON WHEEL WATER TURBINE 1.0. Introduction 1.1.

Introduction to Fluid Machines Fluid machines are devices that transfer energy to or from a fluid. The major classes of machines that add energy to a fluid are (i) pumps (liquids), (ii) fans (gases) and (iii) compressors gases). There are many types of pumps such as centrifugal pumps, piston pumps, vane pumps and diaphragm pumps to name but a few. Turbines extract energy from a fluid. There are also a number of types of turbines ranging from simple devices such as the undershot water wheel and the Pelton wheel to more complex machines such as axial, radial and mixed flow turbines. 1.2. Power and Efficiency Whereas all machines transfer energy, it is the rate of energy transfer (Joules/s or J/s) referred to as the power P (Watts or W) that is of real importance. The efficiency n of a fluid machine is a measure of its performance, that is its ability to transfer as much of the energy that is available to it. It is written as a fraction or as a percentage and is the ratio of the output power to the input power. Power and efficiency are therefore the most important parameters in describing a fluid machine. Normally energy transfer takes place between a mechanical device (such as an electric motor or generator) and a fluid. Consider a centrifugal pump as an example, Figure 1. An electric motor continually supplies energy to the pump via a rotating shaft. The pump, through the interaction of a rotating impeller with the fluid, in doing work on the fluid, transfers energy to the fluid. The power supplied to the pump can be calculated by measuring (i) the torque T (Nm) within the shaft linking the motor to the pump and (ii) the rotational speed of the shaft. The available power is then calculated: P(W) = T(Nm). w(rad/s) Eq. 1 Discharge Created by Malachy O'Rourke Impeller Eye Inflow Page 15 of 23 Blade Casing, housing, or volute Figure 1- A Centrifugal Pump Hub plate 4:04 PM 10/09/2009 Energy stored within a fluid manifests itself as fluid pressure, fluid kinetic energy or fluid potential energy. When energy is transferred to or from a fluid there is an associated change in one of these properties. For example, in a centrifugal pump the fluid exits the pump with an increased pressure. Turbines in general consist of a nozzle that remains stationary relative to a moving rotor. A turbine is classified according to the drop in fluid pressure that occurs across the rotor. When this pressure drop is zero the turbine is known as an impulse turbine otherwise it is known as a reaction turbine. Nozzle vanes MEEN 20010 – Mechanics of Fluids I – Laboratory No. 2 – Analysis of a Pelton Wheel Water Turbine Nozzle ● ● Rotor (a) Pelton Wheel Turbine Bucket Created by Malachy O'Rourke Figure 2 - An Impulse and a Reaction Turbine 1.3. The Pelton Wheel Hydraulic (Water) Turbine The Pelton wheel turbine was first proposed and developed by Lester Pelton (1829-1908), see Figure 2(a). A nozzle is used to accelerate a supply of pressurized water into a high velocity jet. The jet is directed so that it impinges on to the buckets of the Pelton wheel and in so doing applies a force on the buckets and hence a torque on the shaft of the wheel. The power developed within the shaft (and hence the power output of the turbine) is calculated by Eq. 1. In the Pelton wheel turbine the pressure of the working fluid changes only as it moves through the nozzle. Outside of the nozzle the pressure of the fluid is at ambient pressure (normally atmospheric) and no further change takes place. The Pelton wheel turbine is therefore an impulse turbine. (0) (b) Francis Turbine Figure 3- An Industrial Pelton Wheel Turbine 1.4. Laboratory Aims and Objectives The aim of this laboratory is to become familiar with the concept of a fluid machine illustrated through the examination of a Pelton wheel hydraulic turbine. Rotor The objective of this laboratory are: To examine the performance (power and efficiency) of a Pelton wheel turbine as the ratio of the wheel/jet velocity is varied from 0 to 1. To plot the shaft torque, power developed and efficiency of the Pelton wheel turbine. Page 16 of 23 4:04 PM 10/09/2009 MEEN 20010 – Mechanics of Fluids I – Laboratory No. 2 – Analysis of a Pelton Wheel Water Turbine 2.0. Basic Theory The interaction of a jet with a bucket of the Pelton wheel is shown in Figure 4. A jet of velocity V₁ impinges on the bucket at a mean radius rm. The angular velocity of the Pelton wheel is w and the linear velocity at the radius of interaction is U (=w). The direction ab is in the direction of the jet. The direction of the jet is turned through an angle ß relative to the bucket of the Pelton wheel as shown in Figure 5. V1 Hence the power may be written: Created by Malachy O'Rourke The power P developed by the Pelton wheel is given by: 'm V2 Figure 4- Schematic of Pelton Wheel 2.1. Available Power- Power in the Jet V² 2 The energy per unit mass of fluid (J/kg) of a jet moving with velocity V₁ is given by The power of the jet (i.e. the rate at which energy is supplied to the Pelton wheel) is given by the product of the mass flow rate m (kg/s) transported by the jet and the energy per unit mass: 1 Pin=--mV² (W) 2 Tshaft Fabrm (Nm) Radial 2.2. Developed Power Defining Fab as the force exerted on a vane by the jet, in the direction ab, it follows that the shaft torque is given by: Eq. 3 Eq. 4 P = out = Tshaft (W) U Page 17 of 23 Pout = Fabr@=F₁₁₂U (W) m Tangential ·b Eq. 2 Eq. 5 4:04 PM 10/09/2009 MEEN 20010 – Mechanics of Fluids I – Laboratory No. 2 – Analysis of a Pelton Wheel Water Turbine 2.3. Interaction of Jet and Vane a splitter ridge W₁ = V₁ - U W₂ =W₁ = V₁ - U Blade cross section Axial Tangential Figure 5-Control Volume of Bucket b Newton's 2nd Law applied to a rotating system states: b Created by Malachy O'Rourke a Torque T rate of change of angular momentum of the fluid Ve₁ = V₁ = W₁ +U_(m/s) 01 Ve2 = W₂ cos B+U (m/s) 02 Applying Newton's 2nd Law the torque developed in the shaft is given by: Tshaft = mrm (Ve2 - V₁₁) (Nm) 02 01 U - The theoretical torque developed within the shaft is therefore: To determine the rate of change of angular momentum the absolute tangential velocities Vei and V02 are required. From Figure 6 the tangential velocity at inlet and outlet respectively are given by: Ve2 - Vel = (U-V₁)(1-cosß) (m/s) 02 01 Figure 6- Inlet and Outlet Velocity Triangles Tshaft mr (U-V₁)(1-cosß) (Nm) I The efficiency (%) of the Pelton wheel turbine is given by: P out M= x 100 P. in W₁ =W₂ Eq. 8 Neglecting friction, Bernoulli's equation shows that the fluid velocity remains constant through the vane, i.e. W₁ = W₂ = V₁ -U. It follows: U Page 18 of 23 Eq. 6 Eq. 7 b Eq. 10 It follows from Eqs. 4 and 10 and since U = r@ the theoretical power developed by the turbine is given by: m Pout=mU(U-V)(1-cosß) (W) Eq. 11 b Eq. 9 Eq. 12 4:04 PM 10/09/2009 2.4. Performance Mapping Torque, power and efficiency may be plotted against the ratio of wheel to jet velocity, i.e. U/V₁, as shown in Figure 7. The following is observed: ● The theoretical wheel velocity will never exceed that of the jet and the actual wheel velocity is less than that of the jet. The maximum torque in the shaft occurs when the wheel is stationary. The minimum torque in the shaft occurs when the wheel is at its maximum speed. The power output is zero when the wheel is stationary. The power output is zero when the wheel is at its maximum speed. The maximum theoretical power output occurs when the wheel rotates at half the jet velocity. The actual power developed is always less than the theoretical power. ● MEEN 20010 – Mechanics of Fluids I – Laboratory No. 2 – Analysis of a Pelton Wheel Water Turbine ● ● Pout 0 We Created by Malachy O'Rourke shaft U = 0.5V₁ max power 0.2 V₁ Ishaft 0 O 0.4 V₁ The maximum theoretical efficiency is given by: Actual torque 0.6 V₁ U=00rm Actual power n=--(1-cosB) Page 19 of 23 0 O Figure 7- Performance Mapping of Pelton Wheel Turbine That the maximum theoretical power developed occurs at a wheel/jet velocity ratio of 0.5 can be proven by determining the turning point on the power curve. This can be achieved by differentiating Pout with respect to U assuming that the mass flow rate, jet velocity and turning angle are a constant for a particular dP U out=0 shows that =0.5. The maximum theoretical power output machine configuration. Setting == du V₁, is given by: Pout = 0.25mV,² (1-cosß) (W) 0.8 V -I shaft 1.0 V₁ Eq. 13 Eq. 14 4:04 PM 10/09/2009