Impulse Water Turbines Pelton

25
1 IMPULSE WATER TURBINES (PELTON TURBINE) Water turbines are used to generate electrical power, about 1/5 of the World‟s electric power is generated in this way. There are two main types of water turbine, Impulse and Reaction. The Pelton is an impulse turbine; the Francis and the Kaplan are reaction turbines. Figure 1shows a reservoir/turbine/generator arrangement. It will become apparent later that this is a Reaction Turbine power plant. FIGURE 1 IMPULSE TURBINE In this section of the work we will study the Pelton Turbine which is the most common type of impulse water turbine. Some of the diagrams below are taken from Wikipedia where you will find a full description of this type of power generating machine. The design of a hydroelectric scheme is a multi-disciplinary project. As Mechanical Engineers our main interest is in the turbine, specifically in the mechanics of the buckets which are clearly shown in Figure 3 of the Pelton Wheel, and in Figure 4 which shows the nozzles that create the high velocity water jets that turn the wheel. Figure 2 shows a Pelton Wheel installation. This type of turbine is most suited to high head, low water volume conditions. The wheel runs in atmospheric air. The jets when they leave the nozzles are at high velocity and atmospheric pressure. The water when it leaves the blades (buckets) is at atmospheric pressure i.e. in an impulse turbine there is no pressure drop across the blades. The spent water falls into the tailrace. There is no hydraulic connection between the runner and the tailrace. The conditions that exist in a reaction turbine will not be dealt with at this time.

Transcript of Impulse Water Turbines Pelton

Page 1: Impulse Water Turbines Pelton

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IMPULSE WATER TURBINES

(PELTON TURBINE)

Water turbines are used to generate electrical power, about 1/5 of the World‟s electric power is

generated in this way.

There are two main types of water turbine, Impulse and Reaction. The Pelton is an impulse turbine;

the Francis and the Kaplan are reaction turbines.

Figure 1shows a reservoir/turbine/generator arrangement. It will become apparent later that this is a

Reaction Turbine power plant.

FIGURE 1

IMPULSE TURBINE

In this section of the work we will study the Pelton Turbine which is the most common type of

impulse water turbine. Some of the diagrams below are taken from Wikipedia where you will find a

full description of this type of power generating machine. The design of a hydroelectric scheme is a

multi-disciplinary project. As Mechanical Engineers our main interest is in the turbine, specifically

in the mechanics of the buckets which are clearly shown in Figure 3 of the Pelton Wheel, and in

Figure 4 which shows the nozzles that create the high velocity water jets that turn the wheel.

Figure 2 shows a Pelton Wheel installation. This type of turbine is most suited to high head, low

water volume conditions. The wheel runs in atmospheric air. The jets when they leave the nozzles

are at high velocity and atmospheric pressure. The water when it leaves the blades (buckets) is at

atmospheric pressure i.e. in an impulse turbine there is no pressure drop across the blades. The

spent water falls into the tailrace. There is no hydraulic connection between the runner and the

tailrace. The conditions that exist in a reaction turbine will not be dealt with at this time.

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FIGURE 2

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PELTON WHEEL OR RUNNER

FIGURE 3

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FIGURE 4

PELTON WHEEL SHOWING THE WATER NOZZLES.

Before we can turn our attention to an actual turbine we must cover some preliminary

theories.

WATER NOZZLE

A water nozzle, decreases pressure and increases velocity.

Example 1. The volume flow rate of water through a 40 mm diameter pipe is 0.00377 m3/s. If this

pipe terminates in a 25 mm diameter nozzle, determine

(i) the velocity of the water at the nozzle exit,

(ii) the pressure of the water at entry to the nozzle. Neglect losses.

(i)

ANSsmmnozzleofCSA

smrateflowvolumetric

V NOZZLEEXIT

/.

).(

.

)(

)/(687

02504

003770

22

3

(ii) Apply Bernoulli‟s Equation, between Entry to (1) to Exit from (2) the nozzle

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ANSbarORmNpmkgtakegg

p

smVVVAVAflowofContinuity

g

V

g

V

g

pppandZZ

Zg

V

g

pZ

g

V

g

p

ATMOS

2502499110002

3687

314

040687

4

0250

22

22

2

1

322

1

1

22

1122

2

1

2

21221

2

2

221

2

11

.//).(

/.

..

:

;

:

In practice the high velocity water jet is created by a reservoir or dam, FIGURE 2

Example 2: This simple example illustrates the principle. Determine the velocity of the water at exit

from the nozzle. Neglect losses.

ANSsmV

V

gZV

g

VZ

ZVppp

Zg

V

g

pZ

g

V

g

p

ATMOS

/.

.

)(

;;

:

8510

68192

12

2

00

22

2

2

12

2

21

2121

2

2

221

2

11

JET (1)

(2)

40 mm 25 mm

mmmm

mm

3.6 m

2.4 m

DATUM

(1)

(2)

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We need to study how the force that turns the Pelton Wheel is created. To do this we will

commence by considering a simpler application, i.e. where the vane, blade or bucket is fixed.

THRUST OF A JET

On a vertical fixed plate

Newton’s 1

st Law: A body will continue in its state of rest or of uniform motion in a straight line

unless acted upon by an external force.

The water jet is brought to rest, so according to Newton‟s 1st Law there must be a force acting on

the jet. Obviously this force is created by the fixed plate. We need only consider the force normal to

the plate.

Consider the force F, acting normal to the fixed plate when the velocity of the jet of water, normal

to the plate, is reduced to zero velocity.

The fixed plate exerts a force of F (N) which brings the jet to rest.

Newton’s 3rd

Law: To every action there is an equal and opposite reaction

The Jet force (F) is the Equal but opposite Reaction to the plate force (F).

Newton’s 2nd

Law: The force acting on a body of constant mass is equal to the rate of change of its

linear momentum. (the force equation F = ma comes from this law)

Impulse = change in momentum.

Impulse = impulsive force (N) x time of action of the force (s) = Ft (Ns)

Momentum = m (kg) x V(m/s) = mV (Ns)

)()()(

.

)()/(

)(

2

later) (see .for valuenegative a give will

and Jet, on the acting Force Plate theisit i.e. momentum thechanges that force theis This

momentum. of change of rate the force impulsive The

:However

hen constant t is mass When the

12

12

NVVmF

F

(F)ei

velocityinchangemFskgmt

m

velocityinchangemassimpulse

VVmFt

Fixed Plate V m/s

m kg/s F (N)

. F (N)

Water jet

FIGURE 5

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Example 3. The velocity of a water jet is 35 m/s, and the mass flow rate of water is 7.2 kg/s. The jet

is striking a flat vertical fixed plate. What is the jet force acting on the plate?

NFNVVmF 25235029712 )(.)()(

The reason for the negative value.

+ or – are indicators of Direction

Velocity is a VECTOR quantity, it has magnitude, sense and direction. Velocity changes if:

its magnitude changes; or

its direction changes; or

its magnitude and direction changes.

The diagram, above, shows the jet velocity as V m/s , this implies that is the positive direction.

Therefore velocities and forces are +ve and velocities and forces are –ve.

It is the force exerted by the plate or vane on the water that changes the momentum of the water,

therefore it is this force that the equation calculates, so the –ve value is only indicating that the plate

force is acting opposite to the jet direction.

In this example the force of the plate on the water is 252 N , and the force of the jet on the plate is

252 N ANS

THRUST ON A FIXED CURVE VANE OR BLADE.

In this case the velocity does not change in magnitude, it does however change its direction, so

there must be a force causing this change.

Example 4. A jet of water enters a fixed curved vane with a velocity of 100 m/s, follows the curve

of the vane and leaves with a velocity of 100 m/s. The angle between the jet entering and leaving

the vane is 20o. Determine the force acting on the vane. The mass flow rate of water is 8 kg/s.

Sketch the absolute or “relative to the Earth” velocity vectors. (see later)

“o” is a fixed earth point. Velocity vectors drawn from o are absolute velocities.

Let oa = inlet velocity, and ob = the exit velocity

20o

100 m/s

100m/s

Mass flow rate

= 15 kg/s

FIGURE 6

FIXED

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Resolve the vane inlet and exit velocities; horizontally and vertically .

Take as the positive direction.

I.e the horizontal component of velocity at inlet is 100 cos 10O = 98.48 m/s

The vertical component at entry is 100 sin 10O = 17.36 m/s upwards

The horizontal component of velocity at exit is 100 cos 10O = 98.48 m/s i.e -98.48 m/s

The vertical component at exit is 100 sin 10O = 17.36 m/s upwards

You will observe that there has been no change in velocity in the vertical (Y) direction.

There has been a change in the horizontal (X) direction, from 98.48 m/s to – 98.48 m/s.

The force component in the X direction is

)()(

)(

NVVmF

velocityinchangemF

12

F = 8( -98.48 – 98.48) = -1576 N

In this case, because there in no change of velocity in the Y direction, there is no vertical

component of force.

The resultant force is -1576 N i.e 1576 N ; which as before gives the direction of the vane force

acting on the water jet.

The force of the jet on the vane is 1576 N ANS

Example 5. Repeat the above example, taking as the positive direction. You should get the same

answer.

Example 6. A jet of water enters and leaves a fixed vane as shown in figure 8. Friction reduces the

exit water velocity from 20 m/s to 18.5 m/s. The mass flow rate of the water is 0.5 kg/s.

Determine:

i. the “X” axis change in velocity:

ii. the “Y” axis change in velocity:

iii. the force acting in the direction of the “X” axis;

iv. the force acting in the direction of the “Y” axis.

v. The resultant force of the water jet on the vane.

o

a

V1 cos

V1 sin b

o V2 sin

V2 cos

INLET EXIT FIGURE 7

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X axis velocities: +20 m/s, - 18.5 cos 60O.

Change in X axis velocity = (V2 – V1) = -18.5 cos 60O – 20 = -29.25 m/s ANS

Force acting in X direction = 0.5 kg/s(-29.25) = -14.625 N ANS Or 14.625 N

This is the force of the vane on the water jet.

The X direction force of the water jet on the vane = 14.625 N ANS

The Y direction change in velocity = -18.5 sin 60O. = -16 m/s ANS (i.e downwards)

The vane force acting in the Y direction = 0.5 x -16 m/s = -8 N (i.e downwards)

The jet force acting in the Y direction = 8 N (upwards) ANS

The Resultant Force: Jet on vane.

MOVING VANE or BLADE

Before we can consider the forces acting on a moving vane or blade we must have a firm

understanding of absolute and relative velocities.

ABSOLUTE VELOCITY

If you are an observer, standing still, observing the motion of a body or several bodies. You may

consider yourself as an Earth Point (o), and you will be observing absolute or relative to the Earth

velocities.

Example 7: Draw the absolute velocity vectors for 3 cars. A is travelling N at 30 m/s, B is travelling

E at 20 m/s and C is travelling W at 25 m/s.

Absolute velocity vectors always commence at an earth point o. Use lower case for velocity vectors.

20 m/s

18.5 m/s

120o

FIXED

+ Y

+ X

18.5

20

60O

o

FIGURE 8

8

14.625

R = 16.7 Resultant force of jet on vane

= 16.7 N at ANS

FIGURE 9

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RELATIVE VELOCITY

Before we discuss why an understanding of relative velocity is required, we will concentrate on

how to determine the velocity of one body relative to another.

(Do not try to visualise the velocity of one body relative to another, let vector analysis solve the

problem).

Example 8.

(i) Determine the velocity of car B relative to car A.

(ii) Determine the velocity of car C relative to car A.

We imagine that the observer is in car A. It is a simple matter to draw velocity vectors when the

observer is at rest, so let us bring the observer to rest.

We must establish velocity balance, by adding the –ve value of A to B and C. The results are the

velocity of B relative to A and the velocity of C relative to A.

o o b c 20 m/s 25 m/s

o

a

30 m/s

-30 m/s

-30 m/s -30 m/s

B rel to A C rel to A

FIGURE 11

o

o o

a

b c 30 m/s 20 m/s 25 m/s

FIGURE 10

-30 m/s

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Vector addition is when we add vectors, e.g. when determining a resultant force. In this case we are

subtracting one vector from another, i.e. the velocity of one moving body relative to another

moving body is by Vector Difference. There is a simpler way of determining the difference between 2 velocity vectors. Draw the absolute

velocities from a common earth point; then complete the triangle.

The velocity of B relative to A is vector ab

The velocity of A relative to B is vector ba

The velocity of C relative to A is vector ac

The velocity of A relative to C is vector ca

BODIES COLLIDING.

Consider bodies A and B only. This diagram shows the position of the bodies with regard to each

other, i.e. in terms of distance apart and angular relationship. Three supposed position of B are

shown. The path of B relative to A is also shown. Remember, when we show B relative to A, we

can assume A is fixed.

From the displacement diagram you can see that the path of relative velocity, B relative to A must

pass through A for collision to occur. The relative velocity paths of B relative to A drawn from B1

and B3, miss A.

This analysis is the basis of collision studies, on land, sea or in the air.

What significance has it for us?

o c 25 m/s

o

a

b

30 m/s

20 m/s

a

30 m/s

36 m/s 39 m/s

39.8O

FIGURE 12

A

B1

B2

B3

33.6O

DISPLACEMENT DIAGRAM

FIGURE 13

33.6 O

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In our case, to create a force which will move the blades, the high velocity water leaving the nozzle

MUST collide with the moving blades, but this collision must occur without SHOCK.

IMPULSE TURBINES with a blade profile as shown (steam turbines and some water

turbines)

High velocity water leaves the nozzle, which is at angle to the horizontal. If the blade is at rest

then this is the angle at which the water jet will impact the blade, but because the blade is moving

the jet will impact at the angle 1, i.e the angle, to the horizontal, of the velocity of the water relative

to the blade, Vri.. 1 depends on VAi, and U. 1 can be determined. A usual design consideration is

that the water jet enters the blade without shock. This will be achieved if the blade inlet angle is 1.

Example 9. With reference to the previous diagram, the velocity of water leaving a nozzle is 60 m/s.

the blade velocity is 30 m/s. the nozzle is inclined at 20O to the horizontal (i.e = 20

O). Determine

the inlet angle of the blade if the water is to enter the blade without shock.

Draw the absolute velocities of the water and the blade; owi and ob. Join the open end to give Vri.

Using the cosine rule.

sm

Vri

CosVUVUVri

o

AiAi

/.

)cos(

)(

433

20603026030

22

22

22

o b

20O

VAi

U

Vri

1

Not To Scale Velocity diagram at inlet

wi

FIGURE 15

Blade velocity

„U‟ m/s

Absolute velocity of

the Water at inlet

„VAi‟ m/s Vri

Velocity of water

relative to blades at

inlet

1

2

Velocity of water

relative to blades at

outlet Vro

FIGURE 14

SPACE DIAGRAM

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Using the sine rule

ANS

VVri Ai

o

O

1

1

37.9

sine sine angles, acute ingcorrespond of theoremby the

6140433

342060

20.

.

.sin

sinsin

TO REVISE YOUR TRIGNOMENTAL RELATIONSHIPS GO TO:- http://www.themathpage.com/atrig/functions-angle.htm

We must investigate what is happening at the blade outlet.

The water enters the blade at the relative velocity, water to blade at inlet (Vri).

The water leaves the blade at the relative velocity, water to blade at outlet, (Vro).

Blade outlet angle. In many cases 2 = 1, we will assume that this is the case here.

Blade friction. In passing over the blade the magnitude of the relative velocity will be reduced by

the effects of friction. In this case we will assume no relative velocity loss due to friction, i.e Vro =

Vri.

Example 10. Using the conditions given in the previous example. Draw the velocity diagram at

outlet, and determine the absolute velocity of the water at outlet (magnitude and direction). The

blade is symmetrical and friction may be neglected. Remember relative velocities cannot be drawn

from an earth point.

2 = 37.9O, Vro = 33.4 m/s. Recall, Vro is the velocity of the water relative to the blade at outlet i.e

vector b wo

ANS; 21.3 m/s and 74.4O

COMBINING THE INLET AND OUTLET DIAGRAMS

Because the blade velocity is common to both the inlet and outlet diagrams it is convenient to draw

the combined diagram.

o b

VAo

U

Vro

NTS Velocity diagram at outlet

wo

FIGURE 16

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TANGENTIAL AND AXIAL FORCES

Recall from the work on fixed vanes. The change in velocity was resolved into 2 components, one

in the X direction and the other in the Y direction, Figures 7 and 9. With reference to Figure 18, the

axes have been relabelled as T and A. T is tangential to the Pitch Circle Diameter (PCD) of the

runner. The PCD is measured at blade mid-height. A is in line with the runner shaft axis. The

tangential and axial forces can be calculated when the velocities have been determined. The force F,

which acts tangentially to the PCD is the force that drives the runner. Systems are designed so as

the axial force is as low as possible. Axial force plays no part in developing power, it creates end

thrust that has to be accommodated in the bearing design.

THE VELOCITY OF WHIRL.

With reference to Figure 17, the VELOCITY OF WHIRL is the sum of the Tangential components

of Vri and Vro. i.e it is the change in velocity of the water between inlet to and outlet from the blade

in the direction of blade movement. If the mass flow rate of water is

.

/ skgm ,

rad/s

F (N) U m/s

PCD

2

Bearing

Runner

(Rotor)

Runner shaft

PCD F

Blade fixed to rotor FIGURE 18

Rotor or

Runner

T

A

o b

VAi

U

Vri

1

NTS COMBINED VELOCITY DIAGRAM

wi wO

2

Vro VAo

VWHIRL

FIGURE 17

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Then from equation

F = rate of change in momentum in the direction of blade movement

= mass flow rate x velocity of whirl.

)(3WHIRLVmF

i.e F acts tangentially to the PCD of the blade rotation circle, and is the force which drives the

blade.

TURBINE RUNNER POWER (this is the power developed by the rotor or runner as

determined by velocity diagram values. This power is often referred to as the DIAGRAM

POWER. This is the power transmitted to the rotor shaft. The actual power output from the

turbine to the generator would have to take into consideration the friction power loss in, for

example, the bearings and gearbox).

sradN

N

beforeasWrFPOWERRUNNER

smrvelocitybladeNOTE

WvelocityBladeFPOWERRUNNER

OR

WrFPOWERRUNNER

NmrFT

WTPOWERRUNNER

/

)(

)/(:

)()(

.

)()(

)(

)()(

60

2

rev/minin speedrotor if

6

5

4

Example 11. From the previous example Vri = 33.4 m/s, and Vro = 33.4 m/s, blade speed = 30 m/s.

1 = 2 = 37.9O. The mass flow rate of water from the nozzle is 60 kg/s. Determine the runner

power output.

VWHIRL = tangential component of Vro + tangential component of Vri

= Vri cos 37.9O + Vro cos 37.9

O

ANSkWWspeedbladeFPowerRunner

NVVmF rori

99494870303160

3160937937

.

).cos.cos(

Example 12. An impulse steam turbine has blades of equal inlet and outlet angles. Steam leaves the

nozzle with a velocity of 270 m/s. The nozzle is inclined at 11O to the direction of blade motion.

The relative velocity at blade exit is 90% of the relative velocity at blade inlet. The mass flow rate

of steam is 9.56 kg/s. The blade speed is 100 m/s. Determine:

i. the velocity of Whirl,

ii. the tangential force acting on the blade, and

iii. the power (runner) output of the turbine.

SOLVE THIS PROBLEM USING THE FOLLOWING METHODS:

1. By drawing a velocity diagram to scale.

2. Mathematically, using the rules of trigonometry.

Either method is acceptable.

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Always draw a space diagram, it need not be as elaborate as Figure 14, see Figure 19. The space

diagram provides an opportunity to sort out the directions of the velocities.

Blade velocity

„U‟ m/s

Absolute velocity of

the Water at inlet

„VAi‟ m/s Vri

Velocity of water

relative to blades at

inlet

1

2

Velocity of water

relative to blades at

outlet Vro

FIGURE 14

VAi

U

Vri

Vro

FIGURE 19 SPACE DIAGRAM

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FIGURE 20 VELOCITY DIAGRAM

(1)

Draw the absolute velocity of the blade “U”, vector o b.

Draw the absolute velocity of the steam at inlet VAi, vector o si

Complete the inlet triangle by drawing the velocity of the steam relative to the blade at inlet, Vri,

vector b si

MEASURE: 1 = 17 O, therefore 2 = 17

O, Vri = 173 m/s therefore Vro = 0.9x173 = 156 m/s

Draw the velocity of the steam relative to the blade at outlet Vro, vector b so

Complete the outlet triangle by drawing the absolute velocity of the steam relative to the blade at

outlet VAo, vector o so.

RECALL: Absolute or velocities relative to the Earth always commence at the Earth point “o”

Relative velocities do not commence or end at an Earth point.

Velocity of Whirl, measure horizontally across the “peaks” VWHIRL = 314 m/s ANS

3. By calculation:

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ANSkWVelocityBladeF

ANSNVmF

VAoVAiOR

ANSsmVro(Vri

smVro

Vri

V

VVri

smVri

CosVUVUVri

WHIRL

O

oo

oo

Ai

Ai

o

O

AiAi

3001002998

29986313569

11

63136148165317317

6155917290

31711

11

9172112701002270100

2

1

1

22

22

Power Runner Turbine

..

)coscos( whirl of velocity The

/..).cos.cos whirl of velocity The

/...

.sin

sin

o b si sine sine angles, acute ingcorrespond of theorem the bysinsin

/.)cos(

)(

1

THE PELTON WHEEL

The diagram represents the interaction between a Pelton Wheel bucket (vane or blade) and a water

jet.

The bucket has a splitter that divides the water jet into two equal streams. The water enters the

bucket in the direction of bucket movement, and is deflected through an angle . For the Pelton

Wheel it is usual to state the deflection angle.

Follow the rules for constructing the velocity diagram.

Draw the absolute velocity of the buckets, U, vector o b.

Draw the absolute velocity of the water jet, VAi, vector o wi.

The velocity of the water relative to the blade at inlet, Vri, is vector b wi.

U VAi

Vro

Vro

FIGURE 21

Vri

Vri

Space Diagram

. o b wi

VWHIRL

Vro

U Vri

VAo

Velocity Diagram

wo

VAi

SPLITTER

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Draw the velocity of the water relative to the bucket at outlet, Vro, i.e. vector b wo

The absolute velocity of the water relative to the blade at outlet is VAo, i.e vector o wo

VWHIRL = Vri + Vro cos(180 – ) (m/s)

The tangential force causing motion = F x VWHIRL (N)

Turbine Runner Power = F x VWHIRL x Bucket velocity

Turbine Runner Power = F x VWHIRL x U (W)

Example 13. A Pelton turbine has a blade speed of 14 m/s. The mass flow rate of water through the

turbine is 700 kg/s. The effective head of the water supply above the jet is 30 m. The blades deflect

the jet through 160O. There is no loss in relative velocity over the blade. Determine the runner

power output of the turbine.

We need to determine the velocity of the water leaving the jet i.e. VAi

From Equation (1)

smV

smgZV

Ai

Ai

/.).(

/)(

2624308192

(m) head effective Zwere

2

With reference to Figure 21.

VWHIRL = Vri + Vro cos(180 – ) (m/s)

= (24.26 - 14) + ( 24.26 - 14) cos 20O. = 19.9 m/s

From Equations (3) & (4)

ANSkWWPowerRunnerTurbine

speedbladeVmPowerRunnerTurbine

VmF

WHIRL

WHIRL

19519500014919700 )(.

THE RUNNER EFFICIENCY OF A WATER TURBINE ( R)

)()(.

)()(

82

1

7

2 WVmJETOFOUTPUTEKOFRATEJETTHEOFOUTPUTPOWER

JETTHEOFOUTPUTPOWER

OUTPUTPOWERRUNNERTURBINEEFFICIENCYRUNNER

Ai

R

Example 14. Continuing the previous example. What is the efficiency of the turbine runner?

ANSJETTHEOFOUTPUTPOWER

OUTPUTPOWERRUNNERTURBINEEFFICIENCYRUNNER %..

.

6949460

26247002

1

195000

2

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RUNNER MAXIMUM EFFICIENCY

This depends upon the relationship between the blade speed (U) and the velocity of the water at exit

from the nozzle, (VAi).

With reference to the velocity diagram in Figure 21.

FOR THIS ANALYSIS WE MUST USE VECTOR NOTATION.

Recall that F = mass flow rate (change in velocity of the water in direction of the blade movement)

i.e )(.

3WHIRLVmF

VWHIRL =

In this case = ( Vrocos – Vri)

therefore F = mass flow rate( Vrocos – Vri), you will recall from earlier work that this equation

will give the force of the blade on the jet, i.e a –ve value.

However the tangential force of the jet on the blade is )cos( VroVriF

From the velocity diagram:

. o b wi

VWHIRL

Vro

U Vri

VAo

Velocity Diagram

wo

VAi

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21

0.48.about is ratio the windage, and friction bladefor allowing practice, In

)(.U

when maximum a isrunner the of Efficiency)()cos)((

)cos)((

)cos)((

(9) equation Rewrite

t practice constantIn are and speed. blade the torespect with (9) equation ateDifferenti

22 Figure -See-------0 is graph ratio speedspeed/jet blade v efficiencyrunner the of slope the

when maximum a be will efficiencyrunner The

maximum. a is whichfor ratio U

a be will there and betweenthat means This

0 blade the strikenot willjet the if

0 movement blade no is there If

)(]cos)()[(

)(

)(

]cos)()[(

cos)()(

then friction blade to due velocity relative in loss no is there if

1050

020

2

1

12

2

1

1

2

1

1

0

0

9

2

1

2

2

2

2

2

Ai

Ai

Ai

Ai

Ai

AiR

Ai

Ai

R

Ai

Ai

Ai

RAi

R

Ai

AiAiR

R

AiAi

AiAi

Ai

V

UV

V

UV

dU

d

V

UUV

V

UVU

V

VVU U

VU

U

Vm

UVUVmUEFFICIENCY

JETTHEOFOUTPUTPOWER

OUTPUTPOWERTURBINEEFFICIENCY

UVUVmUPowerRunnerTurbine

UVUVmF

VriVroUVVri

Page 22: Impulse Water Turbines Pelton

22

MECHANICAL EFFICIENCY OF THE TURBINE ( M)

POWERRUNNERTUBINE

TURBINETHEOFOUTPUTPOWERACTUALTHEEFFICIENCYMECHANICAL M )(

The actual power output from the turbine to the generator takes into consideration the friction

power loss in, for example, the bearings and gearbox).

NOZZLE EFFICIENCY ( N)

Let Z (m) = Effective head immediately before the nozzle. gZminletnozzleatPowerWater

Let V (m/s) = the water velocity at exit from the nozzle. 2

2VmexitnozzleatPowerWater

If there are losses in the nozzle, the water power at nozzle exit is less than the water power at nozzle

inlet.

gZ

V

gZm

Vm

INLETNOZZLEATPOWERWATER

EXITNOZZLEATPOWERWATEREFFICIENCYNOZZLE N

2

22

2

)(

Blade speed-jet speed ratio U/VAi 0 1

Maximum

Efficiency of runner

FIGURE 22

Page 23: Impulse Water Turbines Pelton

23

OVERALL EFFICIENCY OF THE TURBINE ( O)

INLETNOZZLEATPOWERWATER

INLETNOZZLEATPOWERWATERTURBINEWATEROFEFFICIENCYOVERALL

INLETNOZZLEATPOWERWATER

TURBINEOFOUTPUTPOWERACTUALTURBINEWATEROFEFFICIENCYOVERALL

MRN

MRNO

Example 15. A Pelton turbine has an actual power output of 2 MW. The mechanical efficiency is

80%. The mean diameter of the buckets is 1.5 m, and the runner rotates at 1500 rev/min. The jet

velocity is 1.8 times the bucket velocity. Blade friction factor is 0.97 and the angle of deflection is

165O. Determine;

i. the bucket velocity,

ii. the velocity of the water relative to the bucket at inlet,

iii. the velocity of whirl,

iv. the runner power,

v. the water mass flow rate.

vi. the runner efficiency, and

vii. the overall efficiency assuming no loss at the nozzle,

ANSsmbucketsofradiusMeanN

UvelocityBucket /.

.

)( 7511760

2

5115002

60

2

The absolute velocity of the water at inlet (VAi) = 1.8 x 117.75 = 212 m/s

The velocity of water relative to bucket at inlet = VAi – U = 94.2 m/s ANS

The velocity of the water relative to the bucket at outlet = 0.97 x 94.2 = 91.4 m/s

The velocity of whirl = VWHIRL = Vri + Vro cos (180-165)O = 94.2 x 91.4 cos 15

O =

182.5 m/s ANS

U VAi

Vro

Vro

FIGURE 21

Vri

Vri

Space Diagram

. o b wi

VWHIRL

Vro

U Vri

VAo

Velocity Diagram

wo

VAi

SPLITTER

Page 24: Impulse Water Turbines Pelton

24

The Runner Power:

ANSskgmmVUmrRunnerPowe

ANSMWMW

POWERRUNNERTURBINEPOWERRUNNERTURBINE

MW

POWERRUNNERTURBINE

TURBINETHEOFOUTPUTPOWERACTUALTHEEFFICIENCYMECHANICAL

WHIRL

M

/....

..

.

)(

31165182751171052

5280

2280

6

ANSJETTHEOFOUTPUTPOWER

OUTPUTPOWERRUNNERTURBINEEFFICIENCYRUNNER %..

.

.7959570

21231162

1

1052

2

6

ANSEFFICIENCYOVERALL MRNo %....)( 67676608095701

RUNNER MAXIMUM POWER

This depends upon the relationship between the blade speed (U) and the velocity of the water at exit

from the nozzle, (VAi).

The analysis follows the same form as for maximum efficiency.

A graph of Runner power v Blade speed-Jet speed ratio is of the same shape. The analysis will

result in the same ideal relationship.

For Maximum Power 50.AiV

U in practice the ratio is approximately 0.48

RUNNER SPEED CONTROL Spear (or needle valve)

FIGURE 22

Page 25: Impulse Water Turbines Pelton

25

FIGURE 23

http://www.csus.edu/indiv/h/hollandm/ce135/Photos/PWPhotos.htm

Figure 22 shows a spear (needle valve), and Figure 23 shows the runner and needle valve, the

runner is at rest. The dark section at the bottom of the glass is where the water drops out of the unit.

The drive shaft is also shown.

It is important, especially when generating electrical power, that irrespective of the load, the speed

of the runner remains constant. The power output of the jet is a function of the mass flow rate and

the jet velocity. 2

2VmexitnozzleatPowerWater . When the power requirement of the load

varies the jet power must also vary. I.e. vary either Vorm . However, the blade velocity/jet

velocity determines the efficiency of the runner, so the runner output is varied by varying m . This

is achieved by the design of the needle valve, which maintains a constant jet velocity, at a variable

mass flow rate.