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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.
2
FIGURE 2
3
PELTON WHEEL OR RUNNER
FIGURE 3
4
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
5
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)
6
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
7
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
8
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
9
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
10
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
11
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
12
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
13
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
14
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
15
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.
16
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
17
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:
18
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
19
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
20
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
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
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
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
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
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.