Lab Guide - Pelton Turbine

Performance Test of Pelton Turbine Experiment # 1

I. Objective 1. To perform the Pelton turbine experiment using the theoretical procedures. 2. To acquire the characteristic curve of the Pelton turbine experiment. 3. To test the performance of the Pelton Turbine by varying the parameters.

II. Materials and Apparatus 1 set of Counterweights 1 pc. of Meter stick/ Steel tape 1 set of Pelton turbine set up

III. Theory The Pelton turbine is a hydraulic impulse machine developed in 1889 by Lester Allan Pelton. The buckets are impacted by water jets, and the kinetic energy of the water is transformed into mechanical energy, and finally electrical power. Design of the bucket is an important issue for the turbine efficiency. Pelton turbines are usually used with high heads and with relatively low-volume flows. The water is accelerated through the nozzle and the pressure energy is converted into velocity energy at the outlet of the nozzle. The kinetic energy of the water is converted into rotational energy by deflecting the water jets flow in the impeller, which generates mechanical energy on the shaft.

1 Arioja, Baguio,Bughao DMME USC – TC, Talamaban Cebu Philippines

Figure 1. Pelton turbine parts The runner of the Pelton turbine consists of double hemispherical cups. The jet strikes these cups (buckets) at the central dividing edge which is known as splitter. The water jet strikes edge of the splitter symmetrically and equally distributed into the two halves of hemispherical bucket.

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Figure 2. Runner of Pelton turbine The angular deflection of the jet in the bucket is limited to about (165 to 170 degree). The amount of water discharges from the nozzle is regulated by a needle valve provided inside the nozzle as shown in fig. 3.

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Figure 3. Components of nozzle A. Velocity Components and Efficiency The overall efficiency will be less than maximum theoretical efficiency because of friction in bearing, windage, backsplashing and non-uniform bucket flow. Assuming there are no losses, all of the hydraulic potential energy is converted into kinetic energy, and the theoretical maximum jet velocity is, Ek = Ep mVj2 = mgH Vj =

Eqn. 1

√ 2 gH

In a single Pelton bucket, where u is the impeller velocity and β is the exit angle of the jet. For a Pelton wheel where buckets keep entering the jet and capture all the flow, the mass flow would be ρQ = ρAjVj, from the Euler turbomachine equation. The absolute exit and inlet tangential bucket velocity is equal (u1 = u2 = u). The turbine power relation is then, P = ρQ(u1Vt1 –u2Vt2) = ρQuVj −u[u + (Vj − u)cosβ] P = ρQu(Vj − u)(1 − cosβ) Where u = 2πnr is the bucket linear velocity and r is the pitch radius, or distance to the jet centerline.

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Eqn. 2

Figure 4.: (a) side view of wheel and jet, (b) top view of bucket, (c) velocity diagram The theoretical power of a Pelton turbine is maximum when dP/du = 0 or when, u = 2πnr =

1 V 2 j

Eqn. 3

For a perfect nozzle, the entire available head would be converted to jet velocity Vj = A velocity coefficient Cv is used (Vj = Cv

√ 2 gH .

√ 2 gH ). The theoretical impulse turbine efficiency

than becomes, η = 2(1−cosβ)φ(Cv −φ)

φ=

u √ 2 gH

= peripheral velocity factor

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Eqn. 4

B. Velocity Triangles Since the angle of entry of the jet is nearly zero, the inlet velocity triangle is a straight line, as shown in Fig. 3.2. If the bucket is brought to rest, then the relative fluid velocity, V1, is given by V1 = jet velocity – bucket speed

Eqn. 5

V1 = C1 – U1

Figure 5. Velocity Triangle for Pelton turbine velocity, C2, at exit can be obtained by adding bucket speed vector U2 and relative velocity, V2, at exit. Now using Euler’s equation W = U1CW1 – U2CW2

Eqn. 6

Since in this case CW2 is in the negative x direction, U + V 1 )+ [ V 1 cos (180−α )−U ]} W = U {(

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Eqn. 7

Neglecting loss due to friction across the bucket surface, that is, V1 = V2, then W = U(V1 – V1cos α ¿

Eqn. 8

Therefore, E = U(C1 – U)(1 - cos α ¿ /g

Eqn. 9

By differentiating Eqn. 9 with respect to U and equating it to zero we get, dE (1−cos α ) ( C 1−2 U ) = =0 dU g Then, C1 = 2U

Eqn. 10

E = C12(1 – cos α )/4g

Eqn. 11

Substituting Eqn. 10 to Eqn. 9 we get,

If surface friction is present, V1 ≠ V2, so Eqn. 9 becomes, E = U(C1 – U)(1 - kcos α ¿ /g

where, k =

V2 V1

Introducing hydraulic efficiency as

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Eqn. 12

η=

energy transferred energy available ∈ jet

C. Rectangular Weir Consider a rectangular weir over which the water is flowing as shown in figure 6.

Fig 6 Rectangular Weir

Fig.7 Cross-section view of a rectangular weir The equation is derived in the same way as that for a large rectangular orifice, this simply being the situation where the water surface has fallen below the top of the opening. The basis of

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the method is to apply the energy equation (Bernoulli) to two points on a streamline, point 1 being on the water surface some distance upstream of the weir, and point 2 being in the nappe as it passes over the weir crest at a depth, h, below the water surface. There are a number of assumptions that should be listed in connection with the derivation, since they are of significance later on. They are: i

That the water discharges over the weir from the surface of a large reservoir, so it can be assumed that the velocity of approach is negligible and the pressure is atmospheric. In other words, V1 = 0 and P1 = 0.

ii

That the nappe is at atmospheric pressure. Thus if atmospheric pressure is used as a datum, P2 = 0.

iii

There are no energy losses.

iv

The velocity in the nappe varies with depth, H, that is V = (2gH) 1/2 but there is no variation in velocity across the length, L, of the weir crest.

v

The nappe is as wide as the weir crest, which is it also has a length, L.

vi

The streamlines are horizontal as they pass over the weir crest.

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If a thin horizontal strip of length, L, and thickness, dh, is taken across the nappe at a depth, H, see diagram above, then area of the strip is dA = Ldh; velocity of flow through the strip = (2gh)½; and discharge through the strip, dQ = area x velocity =L(2gh)1/2dh dh at a depth

Let us consider a horizontal strip of water of thickness

h

from the water

surface as shown in the figure; H is the height of the water above the crest of the weir, L is the length of the weir and Cdis the coefficient of discharge. If the area of the strip = we know that the theoretical velocity of water through the strip = through the strip is

√ ( 2 gH )

L. d , and

so discharge

dq=C d ( area of strip )( theoretical velocity ) , substitute the equation we

get: dq=C d ( L . d ) ( √ ( 2 gH ) ) Eqn. 13

The total discharge, over the weir, may be found out by integrating the above equation within the limits 0 and H. 3

2 Q= C d L √ 2 g ( H ) 2 Eqn. 14 3

D. Output Power Measurements of the Pelton Turbine A dynamometer is a device that measures force. In one common type of dynamometer, the force is measured by braking action. One of the first such dynamometers was developed by the French mathematician, Gaspard de Prony (1755-1839) and is called a Prony Brake. The principle of the Prony Brake is that an engine or motor is directly coupled to a drum that has a tensioned, friction belt around it. As the drum revolves, the frictional force is measured.

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Torque is the product of the tangential force and perpendicular distance from the line of action of force of the axis of rotation where F is the tangential force applied and d is the perpendicular distance from the line of action where T =Fd . The power P = Fu or Tω. In terms of u, P = 2ρu(V - u)Q. By taking the derivative of P with respect to u and setting it equal to zero, we find that maximum power occurs when u = V/2, and this power is ρV2Q/2, or ρghQ. This is the energy content of the water from the jet, so the efficiency is unity, with all the energy of the jet turned into shaft output. For any velocity u, the efficiency is η = 4u(V-u)/V2. It is zero for u = 0 and for u = V. This analysis should have been clear and easy to follow. It illustrates the princple of the Pelton wheel very well, and actual wheels are not too far from ideal. When a Pelton wheel is working close to maximum efficiency, the water drops easily from the wheel, with a little turmoil, but not much velocity.

Figure 8. Prony Brake Set up E. Components of the Pelton Turbine

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Figure 9. Components of the Pelton Turbine These are the components of the Pelton Turbine Set up as shown in figure 9. 

Nozzle: – the amount of water striking the vanes (buckets) of the runner is controlled by providing a spear (flow regulating arrangement) in the nozzle.



Spear: – the spear is a conical needle which is operated either by a hand wheel or automatically in an axial direction depending upon the size of the unit.



Runner with bucket: – runner of Pelton wheel consists of a circular disc on the periphery of which a number of buckets evenly spaced are fixed.



Casing: – casing is to prevent the splashing of the water and to discharge water to tail race. It is made up of cast iron or steel plate.



Breaking jet: – when the nozzle is completely closed by moving the spear in the forward direction the amount of water striking the runner reduce to zero.

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But the runner due to inertia goes on revolving for a long time. To stop the runner in a short time, a small nozzle is providing which directs the jet of water on the back of vanes. This jet of water is called breaking jet. 

Governing mechanism: – speed of turbine runner is required to be maintained constant so that electric generator coupled directly to turbine.

F. Working and Efficiency of the Pelton Turbine The amount of water striking the vanes (buckets) of the runner is controlled by providing a spear (flow regulating arrangement) in the nozzle. Then the efficient nozzle that converts the hydraulic energy into a high speed jet. The turbine rotor is called runner. The impact jet of water is striking on the runner and runner revolves at constant with the help of governing mechanism. The runner shaft is connected with the generator; thus the electricity is produce with the help of generator.

The following are the efficiencies of the Pelton Turbine:  

Mechanical efficiencies: – It is ratio of the shaft power to the water power. Hydraulic efficiencies: – It is ratio of the power developed at the turbine runner to the

 

power supplied by the water jet at entrance to the turbine. Volumetric efficiencies: – It is ratio of the theoretical to the actual discharge. Overall efficiencies: – It is ratio of the shaft power to the water power.

IV. Procedure A. Pelton turbine set-up

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Figure 10. Schematic Diagram of Pelton Turbine Set-up 1. Axial pump

Fig.11. Axial Pump

An axial-flow pump, or AFP, is a common type of pump that essentially consists of a propeller (an axial impeller) in a pipe. The propeller can be driven directly by a sealed motor in the pipe or by electric motor or petrol/diesel engines mounted to the pipe from the outside or by a right-angle drive shaft that pierces the pipe. 2. Centrifugal pump

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Fig 12. Centrifugal Pump

A centrifugal pump converts the input power to kinetic energy in the liquid by accelerating the liquid by a revolving device - an impeller. The most common type is the volute pump. Fluid enters the pump through the eye of the impeller which rotates at high speed. The fluid is accelerated radially outward from the pump chasing. A vacuum is created at the impellers eye that continuously draws more fluid into the pump.

3. Pressure tank

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Fig. 13. Pressure Tank A pressure vessel is a closed container designed to hold gases or liquids at a pressure substantially different from the ambient pressure.

Note: When closing the valves, rotate the valve to clockwise direction. Counterclockwise when opening the valve. Run the Pelton Turbine 1. Close the gate valves that may lead to other machineries to ensure that the water is directed to the Pelton turbine,

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2. Make sure that the gate valve near the Pelton turbine is close to protect the turbine buckets form sudden inflows. See fig. 14 below.

Gate Valve

Fig. 14. Gate valve near Pelton turbine 3. The by-pass valve near the centrifugal pump must be fully open and the gate valve near the centrifugal pump must be closed. See fig. 15.

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By-pass Valve

Gate Valve

Fig. 15. Gate valve and By-pass valve near Centrifugal pump

4. Open the primer cup. See fig. 16 below.

Priming Cup Gas cut

Fig. 16. Priming cup and Gas cut

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5. Turn on the axial pump to start priming (See fig. 17). The water is starting to fill the casing of the centrifugal pump. Make sure that the foot valve must be open to ensure the water would pass to the casing of the centrifugal pump. This process may take few minutes.

On switch Off switch

Fig. 17. Axial pump switch 6. After the primer cup is filled with water, make sure that there is no air that is trapped inside the centrifugal pump by opening the gas cut (See fig. 16). Do this when no more bubbles and sound is produced in the primer cup. 7. After ensuring that there is no more air that is being trapped inside, close the gas cut and primer cup. 8. Next step is to turn on the centrifugal pump (See fig. 18). Two students are required for this step. One student will press the switch of the centrifugal pump and the other student will open the gate valve that is near the centrifugal pump as fast as he can. The student who will press the switch will signal the student who open the gate valve on the time the switch was being pressed.

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On switch Off switch

Fig. 18. Centrifugal Pump switch NOTE: Read the instructions carefully and ask questions if not sure 9. After opening the gate valve near the centrifugal pump, open the gate valve near the Pelton turbine slowly until the runner of the Pelton turbine starts to rotate. 10. To control the pressure passing through the gate valve near the centrifugal pump, slowly close the by-pass valve until the desired pressure is achieved.

Start the Experiment proper 1. Adjust the spear setting on the Pelton turbine by rotating the spear rod. Adjust the pressure by turning the by-pass valve slowly. 2. Place the weights on the holder that is connected to the lever of the prony brake. Make sure that the lever is balanced. Record the weight and the speed of the shaft. Calculate the torque. 3. Record the weir height and calculate the flow rate. NOTE: Call the laboratory assistant or instructor if in doubt. Do not run the prony brake without the water lubricating it.

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Shut down the Pelton Turbine 1. Close the gate valve near the Pelton turbine and fully open the by-pass valve. 2. Slowly close the gate calve near the centrifugal pump and when pressure almost reaches to 4 kg/cm2 press the switch of the centrifugal pump and close the gate valve quickly.

V. Suggested Tabulation of Results The results can be presented as a table of results including turbine speed, torque, power, pressure and flowrate. Graphs of spear setting and input and output power may also be presented. Do the procedures as many trials as possible at different pressure.

Table 1. Sample of tabulated data gathered Spear Setting

Load (g)

RPM

Head (cm)

1 2

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Pout (W)

Pin (W)

3 4 5 6 7 8

VI. References Aung Kyaw Minn, (2014), “Design of 225kW Pelton Turbine” DraughtyM, Hydraulic Turbines,McGraw Hill Book Company,1920, Third Edition Garshelis, I. ( 2000). "Torque and Power Measurement." CRC Press LLC

Kjartan Furnes, (2013), “Flow in Pelton Trubines” Marcel Dekker, (2003), “Hydraulic Turbine” R. L. Daugherty and J. B. Franzini, Fluid Mechanics, 6th ed. (New York: McGraw-Hill, 1965). Thurston, R. (1890). “A handbook of engine and boiler trials, and of the indicator and Prony brake: For engineers and technical schools” J. Wiley and Sons. Singal, R. K. (2009). “Hydraulic Machines: Fluid Machinery.” I. K. International Pvt Ltd

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