pelton turbine (fluid machinery)

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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



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.



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:



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.



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



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



Maximum hydraulic efficiency:

taking cos =1, i.e. =

= 1 orl00%.

Q.9. Sketch a pelton turbine bucket and show

its working proportions.

Ans.



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?



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



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



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.



=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 =



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



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,



(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=



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



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



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



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



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, =



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



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, =



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



= 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



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



= 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



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



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.



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



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 =



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



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



=

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.



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



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,



= 0.9279 = 92.79%

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