Chapter 2 _ Pelton Turbine _ Fluid Machinery
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Transcript of Chapter 2 _ Pelton Turbine _ Fluid Machinery
1/27/13 Chapter 2 : Pelton Turbine | Fluid Machinery
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Fluid Machinery
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CHAPTERS
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Topics
Chapter 1 : General Concepts
Chapter 2 : Pelton Turbine
Chapter 3 : Francis and Kaplan
Turbine
Chapter 4 : Centrifugal Pumps
Chapter 5 : Similarity Relations
and Performance
Characteristics
Chapter 6 : Reciprocating
Pumps
Chapter 7 : Hydraulic devices
and Systems
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Chapter 2 : Pelton Turbine
Q. 1. Classify Hydraulic turbine.
Ans. According to the type of energy at inlet
(a) Impulse turbine
(b) Reaction turbine.
According to the direction of flow through
runner
(a) Tangential flow turbine
(b) Radial flow turbine
(c) Axial flow turbine
(d) Mixed flow turbine.
According to the head at inlet of turbine
(a) High head turbine
(b) Medium head turbine
(c) Low head turbine.
According to the specific speed of turbine
(a) Low specific speed turbine
(b) Medium specific speed turbine
(c) High specific speed turbine.
According to the name of the inventor
(a) Pelton turbine
(b) Francis turbine
(c) Kaplan turbine.
Q. 2. What are the factors to be considered in
deciding for a particular hydro electric project.
Ans. (1) Water availability
(2)Water storage
(3)Head of the water
(4)Distance from load centre
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(5)Access to site
(6)Ground water data
(7)Environment aspects of site selection
(8)Consideration of water pollution effects.
Q. 3. Show that the maximum hydraulic
efficiency of a pelton bucket is 100%.
Ans. V = Absolute velocity, = V - v
Hydraulic efficiency=
Q. 4. Distinguish between impulse turbines and
reaction turbines.
Impulse turbine Reaction
turbine
1. All the available fluid
energy is converted in
kinetic energy. 2.
Blades are in action
only when they are in
the front of the nozzle.
3. Water may be
allowed to enter a part
or whole of the wheel
circumference.
4. The wheel does not
run full and air has free
access to the buckets.
5. Unit is installed
above the tail race.
6. There is no loss when
the flow is regulated.
Only a portion of fluid
energy is converted into
kinetic energy.
Blades are in action all
the time.
Water is admitted over
the circumference of the
wheel.
Water completely fills the
vane passages throughout
the operation of the
turbine.
Unit is kept entirely
submerged in water
below the tail race.
There is always a loss
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when the flow is
regulated.
Q. 5. Draw the performance characteristics
curves for both impulse and reaction turbines
and discuss their nature.
Ans. Head, speed and output are the important
factors for designing a turbine. So it required to know
the operating conditions of the turbine under these
variable factors. Information can be obtained
practically by running the turbine system. The results
are drawn in the form of curves are known as the
characteristic curves.
(i) Main or Constant Head Characteristics: When
the head is maintained constant the speed is varied
by quantity of water flow through the inlet the brake
power is measured. The main characteristics of
Francis turbine are identical to those of Kaplan
turbine the discharge characteristics, however, differ
the following information is obtained:
→For pelton turbine discharge curves are the
horizontal lines.
→For Kaplan turbine discharge curve rises as the
speed increases.
→Power and efficiency curves are parabolic in nature.
→For pelton (impulse) turbine the maximum
efficiency for different gate openings occurs at the
same speed.
→For Francis (reaction) turbines the maximum
efficiency for different gate openings usually occurs at
different speeds.
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Fig. Constant head characteristics for Pelton
and Kaplan turbines
(ii) Operating or constant speed
characteristics: The speed is kept constant,
discharge and head H may vary the brake power P
is measured. Overall efficiency is then calculated.
Results are graphically represented as shown in the
figure:
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Fig. Constant speed curves for a hydraulic
turbine
The following information is collected:
→Kaplan turbine is most efficient at all ranges of the
output.
→Different turbines have the same maximum overall
efficiency of about 85% at full load.
→Propeller turbine gives the poorest performance at
part load.
→The performance of the Kaplan and pelton wheel is
much superior at the low heads and at part load.
(iii) Constant Efficiency curves. These curves are
also called as iso-efficiency curves.
The curves are draws after obtaining the data from
various other curves like versus and versus
.
A curve for the best performance is obtained by
joining the peak points of various iso
of various iso efficiency curves as shown in the figure.
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Fig. Constant efficiency curves for a reaction
turbine
Q. 6. Sketch layout of a typical hydroelectric
power plant and label it.
Ans.
Q. 7. What are the factors to be considered in
deciding for a particular hydro electric project.
Ans.1. Water availability
2. Water storage
3. Head of the water
4. Distance from load centre
5. Access to site
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6. Ground water data
7. Environment aspects of site selection
8. Consideration of water pollution effects.
Q. 8. Show that the maximum hydraulic
efficiency of a pelton bucket is 100%.
Ans. V = Absolute
velocity, = V – v
Hydraulic efficiency
For maximum efficiency
2V -4v =0
Or
It means that velocity of the wheel, for maximum
hydraulic efficiency, should be half of the velocity.
Therefore, maximum W.D/kN of water
(Substituting v = )
(1 + cos ) KN
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Maximum hydraulic efficiency:
taking cos =1, i.e. =
= 1 orl00%.
Q.9. Sketch a pelton turbine bucket and show
its working proportions.
Ans.
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Q. 10. Why the buckets of pelton wheel are
provided with an under-cut? What role does the
splitter play in the pelton turbine?
Ans. In pelton wheel each bucket is divided vertically
into two parts by a splitter that has a sharp edge at
the centre and the buckets look like a double
hemispherical cup.
Bucket of Pelton turbine
The striking Jet of water is divided into two parts by
the splitter and each part of the jet flows side ways
round the smooth inner surface of the bucket and
leaves it with relative velocity almost opposite in
direction to the original jet.
Q. 11. What the function of notch in pelton
turbine?
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Ans. A notch made near the edge of the outler rim of
each bucket is carefully sharpened to ensure a loss-
free entry of the Jet into the buckets i.e. the path of
the jet is not obstructed by the incoming buckets.
Q. 12. What are the materials used for the
buckets of pelton turbine?
Ans. The buckets are the most important part of the
pelton turbine they have to be designed to withstand
the full force of the jet. Thus, they are made of
special bronze or steel alloys with nickel, chromium
or stainless steel.
Q. 13. Explain the various factors which decide
the choice for a particular hydraulic turbine for
a hydraulic power project.
Ans.1. Water Availability—The estimates of the
average quantity of water available should be
prepared on the basis of actual measurement. The
curves or graphs can be plotted between the river
flow and time. These are known as hydrographs and
flow duration curves when the river flow data is
calculated on daily, weekly, monthly and yearly
basis.
2. Water-storage-The output of hydropower plant is
not uniform due to wide variations of rainfall. To have
uniform power output water storage is needed so that
excess flow at certain times may be stored to make it
available at the times of low flow.
3. Head of water—The level of water in the
reservoir for a proposed plant should always be
within limits throughout the year.
4. Distance from load centre-To be economical on
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transmission of electric power, the routes and
distances should be carefully considered because cost
depends upon the route selected for the transmission
line.
5. Access to site—It is always desirable factor to
have a good access to the site of the plant. The
transport facilities must taken into the considerations
6. Ground water data-The underground movement
of water has important effects on the stability of
ground slopes and also on the amount and type of
grounding required to prevent the leakage.
7. Environment aspects of site selection—The
project should be designed on the basis so that it
fulfils the following requirement related with
environment
(i) To assure safe, healthful, productive and culturally
pleasing surroundings.
(ii) To avoid health hazards.
(iii) To preserve important historic, cultural and
natural aspects of the site.
8. Consideration of water pollution-The effects of
polluted water on the power plant is one of the major
considerations in selecting the site of hydraulic power
plant. The effects effect the economy and reliability of
the power plant.
Q. 14. Write short note on governing
mechanism for hydraulic turbines.
Ans. Hydraulic turbines are directly coupled to
alternators which must run continuously at constant
speed, so that electricity is produced at constant
frequency. The power produced by water turbine is
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directly proportional to the available head and
discharge through the turbine. The quantity of water
flowing can be controlled by varying the area of flow
at the turbine inlet.
In pelton turbine, the flow area is
changed by moving the spear inside the nozzle and in
reaction turbine, the area of flow is varied by rotating
the guide vanes with the help of governor in a
controlling unit.
Q. 15. Obtain Hydraulic efficiency and work
done by pelton turbine (Impulse turbine).
Ans.
Fig. Triangle of the velocities
V = Absolute velocity of entering water
= Relative velocity of water
= Velocity of flow at inlet.
= Corresponding values at outlet
D= Diameter of wheel
d= Diameter of the nozzle
N= Revolution 9f the wheel in r.p.m.
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=Angle of blade tip at outlet
H= Total head of water
In case, a=0°, =0°, =v and =v-v
The relation between two velocity triangles is
=v and = (V-v)
Force, KN of water in the direction of motion of jet
Work done = Force x Distance=
=
Hydraulic efficiency,
Consider case in which the value of is negative as
shown in figure.
So, Work done per kN of water =
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Q. 16. Explain the various design aspects of
pelton wheel.
Ans.
1. Velocity of Jet –
Theoretical velocity,
Actual velocity, V
Value, , 0. 97 - 0 .99 (Friction loss)
2. Speed Ratio, — It represents the ratio of the
peripheral velocity to the theoretical velocity of the
jet.
Value of = 0.45 - 0.47
3. Mean diameter of the wheel — D refers to the
diameter of the wheel measured upto the centers of
the buckets. The diameter is calculated from the
formula
U
D u the pitch or mean diameter.
4. Jet ratio — m represents the ratio of the pitch
circle diameter of the jet diameter. i.e. m=D/d
5. Number of jets—Pelton wheel has one nozzle or
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one jet. A number of nozzles may be employed when
more power is required.
6. Working proportions-The working proportion of
the turbine bucket are generally specified in terms of
jet diameter d, and usually adopted values are
Axial width, B = 3d to 4d
Radial length, L = 2d to 3d
Depth, T = 0.8d to 1.2d
7. Number of buckets — No. of buckets are
decided on the following principles:
(i) The number of buckets should be as few as
possible so that there is little loss due to friction.
(ii) The jet of water must be fully utilized so that no
water from the jet goes waste.
Q. 17. Define the term Net or effective head.
Ans. The head available at the entrance to the
turbine is called Net or effective head.
H=
Where is the difference of Head race and tail race.
is the loss in head due to friction in penstock.
(1) Work done by pelton wheel, W =
(2) Efficiency of Pelton wheel,
(3) Maximum,
(4) Gross Head, H =
(5) Power supplied the jet = WQH =
(6) Power delivered by the bucket wheel=
(7) Overall efficiency,
(8) Volumetric efficiency,
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(9) Hydraulic efficiency,
(10) Mechanical efficiency,
Problem 1. A double jet Pelton wheel has a
specified speed of 16 and is required to deliver
1000 kW The supply of water to the turbine is
through a pipeline from a reservoir whose level
is 350 m above the nozzles Allowing 5% for
friction loss in pipe make calculations for speed
in rev/mm, diameter of jets and mean diameter
of bucket circle. Take velocity co-efficient =
0.98, speed ratio = 0.46 and overall efficiency
= 85%.
Ans. No. of Jets= 2
=16
P=l000kW=l000x
W
H = 350 — (0.05 x
350) = 332.5 m
d (Jet dia) =?
N (Speed)?
D (mean dia of bucket circle) =?
= 0.98
Ku = 0.46
= 0.85
Power for single jet = = 500 kW
16=
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N= = 1016 r.p.m
0.85=
Q= = 0.36
For single jet q=
= 0.0538m
Problem 2. A pelton wheel is to be designed for
the following specifications. Shaft power =
11,772 kW, head = 380 meters, speed = 750
r.pm, overall efficiency = 86%, jet diameter is
not exceed one-sixth of the wheel diameter.
Determine:
(1) Wheel diameter
(2) No. of jets required
(3) Diameter of the jet
Solution. Given
Shaft power, S.P. = 11,772 kW
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Head, H =380m
Speed, N = 750 r.p.m.
Overall efficiency, = 86 %, or 0.86
Ratio of jet dia to wheel dia. =
Coefficient of velocity, = 0.985
Speed ratio, = 0.45
Velocity of jet, = 85.05
m/s
The velocity of wheel, u =
= Speed ratio x
= 0.45 x =
38.85 m/s
But U =
38.85 =
Or D =
0.989 M Ans.
But
Dia. of jet, d = = 0.165 m
Discharge of one jet, q = Area of jet x velocity of jet
(0165) x85.05
Now
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0.86
Total discharge, Q =
Number of jets
2jets.
Problem 3. The following data is related to a
pelton wheel:
Head at the base of the nozzle 80 m
Diameter of the jet 100 mm
Discharge of the nozzle =
Power at the shaft = 206 Kw
Power absorbed in mechanical
resistance = 4.5 kW
Determine (i) Power lost in nozzle
(ii) Power lost due to hydraulic
resistance in the runner.
Solution. Given
Head at the base of nozzle, = 80 m
Diameter of jet, d = 100 mm = 0.1 m
Area of the jet, a = = 0.007854
Discharge of the nozzle, Q = 0.30
Shaft power, S.P = 206 kW
Power absorbed in mechanical resistance =
4.5 kW
Now, discharge, Q = area of jet x velocity
of jet
0.30 = 0.007854 x
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Power at the base of the nozzle in kW =
=235.44
Power corresponding to kinetic energy of the jet in
kW
= 218.85 kW
(1) Power at the base of the nozzle =
Power of the
jet + Power lost in Nozzle
235.44= 218.85
+ Power lost in nozzle
Power lost in nozzle =
235.44 — 218.85 = 16.59 kW
(2) Also power at base of nozzle = Power at shaft +
Power lost m nozzle
+ Power lost in
runner
+ Power lost due
to mechanical resistance
235.44=206 + 16.59 + Power lost in runner
+ 45
Power lost in runner = 235.44 — (206 +
16.59 + 4.5)
= 235.44 —
227.09 = 8.35 kW.
Problem 4. A pelton wheel is supplied with
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water under a head of 45m and at a rate of 48
/min. The buckets deflect the jet through
165° and the mean bucket speed 14m/s.
Calculate the power delivered to the shaft and
overall efficiency of the machine. Assume
coefficient of viscosity 0.985 and mechanical
efficiency 0.95.
Solution. Power developed is given by
w = 9810
W =9810x0.8=7848N/s
v = 0.985 = 29.267
m/s
u = 14 m/s
K =1 (Neglecting friction in
buckets)
=180°—165° =150°
=cos15°0.9659
[(29.267— 14) + (1 +
0.9659)] 14
= 800 (15.267 + 1.9659) 14
=193.008 kW
Hydraulic efficiency, =
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Overall efficiency,
= 0.98 x 0.95
= 0.9320
= 93.20%
Problem 5. A pelton wheel is required to
develop 4000 kW at 400 revs/min, operating
under an available head of 350 m. There are
two equal jets and the bucket deflection angle
is 165°. Calculate the bucket pitch circle
diameter, the cross-sectional area of each jet
and the hydraulic efficiency of the turbine.
Make the following assumptions (i) Overall
efficiency is 85%, when the water is
discharged from the wheel in a direction
parallel to the axis of rotation (ii) coefficient of
velocity of nozzle ku 0.97 and blade speed ratio
ku 0.46 (iii) relative velocity of water at exit
from the bucket is 0.86 times the relative
velocity at inlet.
Solution. Power available from the turbine shaft
4000x =
(9810xQx350)x0.85
Total discharge through the wheel,
Q= = 1.37
Velocity of jet, v = kv
= 0.97 x
Total discharge,
1.37 = 2A x 80.38
Area of each jet, A =
Velocity of bucket, u=
= 38.12 m/s
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Also
38.12 =
=1.82m
Since the jet gets deflected 165°, (180 — 165) =
15°
Bucket to jet speed ratio,
Invoking the relation for the hydraulic efficiency of a
pelton wheel,
(relative to K.E. of
jet)
= 2 x (0.474 — 0.4742) (1+ 0.86 cos
15°)
= 0.9128 or 91.28%
(relative to
power)
= 0.858 = 85.8%
Problem 6. At hydroelectric power plant, water
available under a head of 250 m is delivered to
the power house through three pipes each 250
m long. Through three pipes the friction loss is
estimated to be 20 m. The project is required to
produce a total shaft output of 13.25 MW by
installing a number of single jet pelton wheels
whose specific speed is not to exceed 38.5. The
wheel speed is 650 rpm, overall efficiency is
0.85 and speed ratio is 0.46. Determine (i) the
number of pelton wheels to be used (ii) Jet
diameter (in) diameter of supply pipe Take
velocity coefficient for the nozzle and Darcy’s
friction factor as 0.97 and 0.02 respectively.
Solution. Net available head, H = 250 — 20
= 230 m
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We know
Power available, P = = 2829
kW
Number of machines = =
=4.68; say 5
(ii) Velocity of jet, V = = 65.16
m/s
Tangential velocity of bucket,
u = 0.46 x Velocity of jet = 0.46
x 65.16 = 29.97 m/s
also ; 29.97=
Diameter of wheel, D = = 0.881 m
Power available from turbine
P = (w Q H)
= (9810 x Q x 230) x 0.85
Discharge, Q =
Also
1.38=
Hence jet diameter, d = 0.167 m
Total discharge for 5 machines = 5 x 1.38 = 6.9
Discharge per pipe, QP =
Loss of head, =
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20=
=0.933m.
Problem 7. A pelton wheel is required to
generate 3750 kW under an effective head of
400 meters. Find the total flow in liters /second
and size of the jet. Assume generator efficiency
95%, overall efficiency 80%, coefficient of
velocity 0.97, speed ratio 0.46, If the jet ratio
is 10, find the mean diameter of the runner.
Solution. Given P = 375OkW, H = 400m,
=95% = 80%
= 0.97, speed
ratio = 0.46
Total flow of water in liters/second
Overall efficiency, 0.8
Size of jet:
Let d = Diameter of the
jet
We know that velocity of the Jet, v
= 85.9m/s
Total discharge = discharge through the
Jet
Q
1.25 = 85.9 = 67.5
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= 1.25/67.5 = 0.0185
Or d = 0.136m = 136mm.
Problem 8. Design a pelton wheel for a head of
350 m at a speed of 300 r p m Take overall
efficiency of the wheel as 85% and ratio of jet
to the wheel diameter as 1/10
Solution. Given H = 350m, N =
300r.p.m.
1. Diameter of the wheel
And peripheral velocity of the wheel
V
=0.46V=0.46x81.2=37.4m/s
Peripheral velocity (v)
D =37.4/15.7=2.4m
2. Diameter of jet,
=240mm
3. Width of the buckets
= 5 x d = 5 x 0.24 = 1.2m
4. Depth of the buckets
= 1.2 X d = 1.2 x 0.24 =
0.48m
5. No. of buckets
Problem 9. A pelton wheel has a mean bucket
speed of 12 m/s and is supplied with water at a
rate of 750 liters per second under a head of 35
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m. If the bucket deflects the jet through an
angle of 160°, find the power developed by the
turbine and its hydraulic efficiency. Take the
coefficient of velocity as 0.98. Neglect friction
in the bucket. Also determine the overall
efficiency of the turbine if its mechanical
efficiency is 80%.
Solution. The power developed by the turbine is
given as
W=wQ
W =9810
W = 9810x 0.75 = 7357.5 N/s
=25.68 m/s
u =12 m/s
k =1 (for neglecting the friction in the
buckets)
= (180°— ) = (180 — 160°) = 200
= cos = 0.9397
Thus by substitution, we get
[(25.68 -12) (1+0.9397) x12W
=238816 W= kW=238.816kW
Since I metric h.p. = 735.5 W, the power
developed by the turbine in metric h.p. is
= 324.699
metric h.p.
The hydraulic efficiency of the turbine is given as
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= 0.966 or 96.6%
The overall efficiency of the turbine given by equation
= 80% or 0.80
= 0.966 x 0.80
= 0.773 or 77.3%
Problem 10. The following data were obtained
from a test on a Pelton wheel:
(a) Head at the base of the nozzle = 32 m
(b) Discharge of the nozzle = 018
(C) Area of the jet= 7500 sq mm
(d) Mechanical available at the shaft = 44 kW
(e) Mechanical efficiency = 94%
Calculate the power lost (z) in the nozzle, (ii) in
the runner, (in) in mechanical friction.
Solution. Power at the base of the nozzle =
(9810 x 018 x 32)
=56510W=56.5lkW
Velocity of flow through the nozzle
v=
Power at the nozzle exit (e g, Kinetic energy
of the jet)
= 51840
W = 51 84 Kw
Power lost m the nozzle = (5651 -51 84) = 467
kW
Power supplied to the number is equal to the
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kmetic energy of the Jet
=51.84kW
Power developed by the runner = = 46.81 kW
Power lost in the runner = (51.84 — 46.81) = 5.03
kW
Power lost in mechanical friction = (46.81— 44) =
2.81 kW
As a check on computation, the difference of power at
the base of the nozzle and the power available at the
shaft must be equal to the sum of the power lost in
the nozzle, i the runner and in mechanical friction.
Thus, we have
(56.51—44) =12.51kW
And (4.67+5.03+281)
=12.51kW
Problem 11 How does a single Jet Pelton wheel
differ from a multi-jet wheel A Pelton wheel is
required to develop 6 MW when working under
a head c 300 m it rotates with a speed of 550
rpm assuming jet ratio as 10 and overall
efficiency as 85%, calculate. (i) Diameter of
wheel (ii) quantity of water required and (iii)
number of jets. Assume suitable values for the
velocity coefficient and the speed ratio.
Solution. Velocity of Jet, V =
= 0.98 -.12 x 9.81 x 300
75.18 m/s
(Assuming = 0. 9)
Tangential (peripheral) velocity of wheel,
= 35.29
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m/s (assuming K,, 0.46)
Also
Bucket pitch circle diameter, D=
Diameter of jet, d =
(b) Power available from the turbine shaft is,
P =
Total discharge through the Pelton wheel, Q =
(c) The discharge through the wheel is supplied by
the jets. Thus
Number of jets, n 2.398/0.8856 =
2.70
And hence we employ three jets.
Revised jet diameter follows from the relation,
2.398=
d = 0.1164 m
Problem 12. A Pelton wheel of 12 m mean
bucket diameter works under a head of 650 m.
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The jet deflection is 165° and its relative
velocity is reduced over the buckets by 15%
due to friction. If the water is to leave the
bucket without any whirl, determine: (a)
rotational speed of the wheel, (b) ratio of
bucket speed of jet velocity, (c) impulsive force
and the power developed by the wheel, (d)
available power (water power) and the power
input to buckets, and (e) efficiency of the
wheel with power input to buckets as reference
input. Take = 0.97.
Solution. Velocity of jet
Let bucket speed
Relative velocity at inlet
Relative velocity at outlet,
Since the jet gets deflected through 165°, the blade
angle at exit,
=180—165=15°
As the jet leaves the bucket without any whirl, the
velocity triangle at outlet will be:
From expression (i) and (ii),
0.85 (109.6 - u) cos
15° = u;
89.98-
0.821u=u
Blade speed u =
89.98/1.821 = 49.41 m/s
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49.41 =
Rotational speed of wheel, N =
(b) Ratio of bucket. Speed to jet speed,
=0.4508
(c) Discharge through the wheel
Impulsive force on the buckets,
= 1000 x 0.86 x 109.6 = 94256
N
Power develop by wheel, P = impulsive force x
distance moved =
= 94256 x 46.41 = 4657 x
W = 4657 kW
(d) Available power (water power)
= wQH = 9810 x
0.86 x 650
=5455x
W=5455kW
Power input to buckets =
=5165x W=5165kW
(e) Efficiency of wheel,
Problem 13. The following data relate to a
Pelton wheel : Head = 72 m, speed of wheel =
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240 rpm., shaft power of the wheel = 115 kW,
speed ratio = 0.45, coefficient of velocity 0.98,
overall efficiency = 85%. Design the Pelton
wheel.
Solution. Power available from turbine shaft
= (WQH) x
115. X = 9810
x Q x72 x 0.85
Q =
Velocity of jet, V=
=
36.81 m/s
=
17.28 m/s
Diameter of wheel
17.28 =
D =1.37m
Diameter of Jet Discharge
0.91 =
d =0.081m81 mm
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Size of buckets
Width of buckets
=5xd=5x81=405mm
Depth of buckets = 1.2 x d = 1.2 x
81= 97.2 mm
No. of buckets on the wheel
Z =
Problem 14. Water under a head of 300 m is
available for the hydel-plant situated at a
distance of 235 km from the surface the
frictional losses of energy for transporting
water are equivalent to 26 J/N. A number of
Pelton wheels are to be installed generating a
total output of 18 MW Determine the number of
units to be installed, diameter of the Pelton
wheel and the Jet diameter when the
followings are available, wheel speed 650 rpm,
ratio of bucket to jet speed 0.46, specific speed
not to exceed 30 (m, kW, rpm), Cv and Cd for
the nozzle are 0.97 and 0.94 respectively and
the overall efficiency of the wheel 87%.
Solution. Given:
Total head=300 m
Length = 2.35 km =
2350 m
Frictional losses = 26 (J/N) 26 (Nm/
N) (as J = Nm) 26 m
Net head, H = 300
-26 = 274 m
Total output = 18
MW = 18 x kW
N= 650
r.p.m.
Ratio of bucket to jet speed =0.46
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=
O.97, = 0.94
= 87%
= 0.87
And =30
where H is in m, P in kW and
N in r.p.m.
Find: (i) Number of units to be installed
(ii) Dia. of Pelton wheel (D)
(iii) Dia. of jet of water (d)
(i) Number of units to be installed
Let P = Power
output of each unit in kW
Using equation (18.28) as
Squaring both sides, we get P =
(ii) Dia. of Pelton wheel (D)
Velocity of jet is given by,
But ratio of bucket to jet speed = 0.46
Or
= 0.46 x = 0.46 x 71.12 = 32.715m/s.
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But
Dia. of Pelton wheel = 0.945 Ans.
(iii) Dia. of jet (d)
We know
Or
Total water power in kW =
Water power in kW per unit =
=2.955x
kW
But water power in kW per unit is given by equation
as,
Water power=
=9.8lxQx274
But discharge (Q) through one unit is also given by
Q=
Or
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Or
d. = = 0.1424 m = 142.4 mm. Ans.
Problem 15. A pelton wheel is supplied with
water under a head of 45m and at a rate of 48
. The buckets deflect the jet through 165°
and the mean bucket speed is 14 m/s.
Calculate the power delivered to shaft and
overall efficiency of the machine. Assume
coefficient of velocity 0985 and mechanical
efficiency 095.
Ans. Given:
H =45 m, Q = 48 min =0.8 /sec
The power developed is given by.
P=
(where k = 0.95)
= 0.8 x 175.952 x
1.9176
=269.92kW
Overall efficiency,
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= 0.9279 = 92.79%