Chapter 2 _ Pelton Turbine _ Fluid Machinery

1/27/13

Chapter 2 : Pelton Tur bine | Flui d Machi ner y

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Chapter 2 : Pelton Turbine

Topics Chapter 1 : General Concepts Chapter 2 : Pelton Turbine Chapter 3 : Franci Chapter Franciss and K aplan Turbine Chapter 4 : Centrifugal Pumps Chapter 5 : Similarity Relations and Performance Characteristics Chapter 6 : Reciprocating Pumps Chapter 7 : Hydraulic devices and Systems

Q. 1. Classify Hydraulic turbine. Ans. According to the type of energy at inlet (a) Impulse I mpulse turbine turbine (b) Reaction turbine. According to the direction of flow through runner (a) Tangential flow turbin turbine e (b) Radial flow turbine (c) Axial flow turbin turbine e (d) Mixed flow turbin turbine. e. According to the the head at inlet of of turbine (a) High head turbine (b) Medium head turbine (c) Low head turbine. 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 the name of the inventor inve ntor (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 Ans (2)Water storage (3)Head of the water (4)Distance (4)Dis tance from load l oad centre

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(5)Access to site (6)Ground (6)Groun d 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.. Ans

V = Abs Absolute olute velocity,

=V-v

Hydraulic efficiency=

Q. 4. Distinguish between impulse turbines and reaction turbines. Impu Im pullse tur urbi bine ne

Rea eact ctiion turbine

1. All the available fluid Only a portion of fluid energy is converted in energy is converted into kinetic Blades

energy. are

in

2. kinetic energy.

action Blades are in action all

only when they are in the time. the front of the nozzle. 3.

Water

may

be

allowed to enter a part or whole of the wheel circumference.

Water is admitted over the circumference of the wheel. Water completely fills the

4. The wheel does not vane passages throughout run full and air has free the operation of the access to the buckets. turbine. 5.

Unit

is

installed Unit is kept entirely above the tail race. submerged in water 6. There is no loss when below the tail race. the flow is regulated. re gulated. ptumech.l or emate.com/fl uid- machi ner y/node/12

There is always a loss 2/38

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(5)Access to site (6)Ground (6)Groun d 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.. Ans

V = Abs Absolute olute velocity,

=V-v

Hydraulic efficiency=

Q. 4. Distinguish between impulse turbines and reaction turbines. Impu Im pullse tur urbi bine ne

Rea eact ctiion turbine

1. All the available fluid Only a portion of fluid energy is converted in energy is converted into kinetic Blades

energy. are

in

2. kinetic energy.

action Blades are in action all

only when they are in the time. the front of the nozzle. 3.

Water

may

be

allowed to enter a part or whole of the wheel circumference.

Water is admitted over the circumference of the wheel. Water completely fills the

4. The wheel does not vane passages throughout run full and air has free the operation of the access to the buckets. turbine. 5.

Unit

is

installed Unit is kept entirely above the tail race. submerged in water 6. There is no loss when below the tail race. the flow is regulated. re gulated. ptumech.l or emate.com/fl uid- machi ner y/node/12

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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 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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Chapter 2 : Pelton Turbine | Fluid Machinery

Fig. Constant head characteristics for Pelton and Kaplan turbines (ii)

Operating

characteristics: discharge

The

or speed

constant is

kept

speed constant,

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 ptumech.loremate.com/fluid-machinery/node/12

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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%. V = Absolute

Ans. 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 ptumech.loremate.com/fluid-machinery/node/12

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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? ptumech.loremate.com/fluid-machinery/node/12

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Ans. A notch made near the edge of the outler rim of each bucket is carefully sharpened to ensure a lossfree 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 ptumech.loremate.com/fluid-machinery/node/12

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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 ptumech.loremate.com/fluid-machinery/node/12

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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 ptumech.loremate.com/fluid-machinery/node/12

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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

(3)

wheel,

Maximum,

(4)

Gross Head, H =

(5)

Power supplied the jet = WQH =

(6)

(7) (8) ptumech.loremate.com/fluid-machinery/node/12

Pelton

Power

delivered

by

the

bucket

wheel=

Overall efficiency, Volumetric efficiency, 15/38

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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%. No. of Jets= 2

Ans.

=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=

For single jet

= 0.36

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 ptumech.loremate.com/fluid-machinery/node/12

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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, Speed ratio,

= 0.985

= 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 ptumech.loremate.com/fluid-machinery/node/12

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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 discharged

is

85%,

from

the

when

the

wheel

in

water a

is

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= Velocity of jet, v = kv

Loremate.com

Like

Related:

6,099

Heat Capacity

= 1.37

|

= Equipment 0.97 x Heat Transfer Search

| Heat And Mass Transfer | Heat Tran

Total discharge, 1.37 = 2A x 80.38 Area of each jet, A = Velocity of bucket, u= = 38.12 m/s ptumech.loremate.com/fluid-machinery/node/12

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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 ptumech.loremate.com/fluid-machinery/node/12

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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, ptumech.loremate.com/fluid-machinery/node/12

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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 =95%

P = 375OkW, H = 400m,

= 80% = 0.97, speed

ratio

= 0.46

Total flow of water in liters/second

Overall efficiency, 0.8

Size of jet: d = Diameter of the

Let jet We

know

that

velocity

of

the

Jet,

v

= 85.9m/s Total discharge = discharge through the Jet Q 1.25 = 85.9 ptumech.loremate.com/fluid-machinery/node/12

= 67.5 25/38

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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 ptumech.loremate.com/fluid-machinery/node/12

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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.

efficiency

of

Also

the

determine

turbine

if

its

the

overall

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 ptumech.loremate.com/fluid-machinery/node/12

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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 ptumech.loremate.com/fluid-machinery/node/12

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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. ptumech.loremate.com/fluid-machinery/node/12

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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. Velocity

Solution.

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; 89.980.821u=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

=5165x

buckets

=

W=5165kW

(e) Efficiency of wheel,

Problem 13. The following data relate to a Pelton wheel : Head = 72 m, speed of wheel = ptumech.loremate.com/fluid-machinery/node/12

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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 ptumech.loremate.com/fluid-machinery/node/12

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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 ptumech.loremate.com/fluid-machinery/node/12

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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

Squaring

both

sides,

(18.28)

we

get

as

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 ptumech.loremate.com/fluid-machinery/node/12

= 0.46 x 71.12 = 32.715m/s. 35/38

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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 ptumech.loremate.com/fluid-machinery/node/12

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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, ptumech.loremate.com/fluid-machinery/node/12

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Chapter 2 _ Pelton Turbine _ Fluid Machinery

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