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.
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Transcript
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