fluid machinery module 1 - ktu notes

Fluid Machinery Module 1

Nishant T

Asst Prof

ISSAT

18-May-17 1 NISHANT, ISSAT

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• A jet of fluid from the outlet of Nozzle flows with high velocity. Hence the kinetic energy of flowing fluid is very high. This stream of fluid from the outlet of nozzle is known as fluid jet.

• In a turbine this fluid jet is made to flow over a series of plane (Vanes), thus exerting some force on the plate. This force is calculated using Impulse-Momentum principle.

• The following cases of impact of fluid jets is important

1. Force exerted by jet on a stationary plate

1. When flat plate is held normal to the jet

2. When flat plate is held inclined to the jet

3. When a curved plate is placed infront of a jet

2. Force exerted by jet on a moving plate

1. When flat plate is held normal to the jet

2. When flat plate is held inclined to the jet

3. When a curved plate is placed infront of a jet


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Force exerted by the jet on a stationary plate held normal to the jet

• Consider a fluid jet from a nozzle which strikes on a flat plate as shown. Let, • V = velocity of jet • d = dia of the jet • The fluid jet after striking the plate get deflected to 90° as shown. Hence the

component of velocity in the direction of jet becomes zero after striking the plate.

• According to newtons law, force exerted is equal to the rate of change in momentum

• Force exerted by the jet on the direction of flow is given by Fx and perpendicular to the direction of flow is given by Fy

• Fx = Rate of change in momentum in the direction of flow = mass striking per sec x change in velocity = ρ a V [ V -0 ] = ρ a V² • Fy = Rate of change in momentum perpendicular to the flow = 0 ( Force exerted is opposite to each other. Hence cancels) • Work done = Force x Distance = ρ a V² x 0 =0 • Efficiency = Output of jet / Input of jet = Workdone per sec/ KE per sec = 0


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Figure


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Force exerted by the jet on a stationary plate held inclined to the jet

• Consider a fluid jet from a nozzle which strikes on a flat plate as shown. Let, • V = velocity of jet • d = dia of the jet • θ = angle between the jet and the plate • Fn = force exerted by the jet on the plate in the direction normal to the plate • Fn = Rate of change in momentum in the direction normal to the plate = mass striking per sec x change in velocity = ρ a V [ V sin θ -0 ] = ρ a V² sin θ • Fn can be be resolved in to two components , Fx which is parallel to the

direction of jet and Fy perpendicular to the direction of jet • Fx = Fn sin θ = ρ a V² sin θ * sin θ = ρ a V² sin² θ • Fy = Fn cos θ = ρ a V² sin θ * cos θ = ρ a V² sin θ cos θ • Work done = Force x Distance = ρ a V² x 0 =0 • Efficiency = Output of jet / Input of jet = Workdone per sec/ KE per sec = 0


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Force exerted by the jet on a stationary curved plate, when jet strikes at the centre

• Consider a fluid jet from a nozzle which strikes on a curved plate as shown. Let,

• V = velocity of jet • d = dia of the jet • θ = angle made by the jet with horizontal after striking the plate

• The fluid jet after striking the plate get deflected as shown and comes out with the same velocity in the tangential direction of the curved plate. According to newtons law, force exerted is equal to the rate of change in momentum

• Fx = Rate of change in momentum in the direction of flow = mass striking per sec x change in velocity = ρ a V [ V – (- V cosθ)] = ρ a V [V +V cosθ] = ρ a V2 [1 + cosθ] • Fy = Rate of change in momentum perpendicular to the flow = 0 ( Force exerted is opposite to each other. Hence cancels) • Work done = Force x Distance = ρ a V² x 0 =0 • Efficiency = Output of jet / Input of jet = Workdone per sec/ KE per sec = 0 • Angle of deflection of the plate = 180- θ


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figure


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Force exerted by the jet on a stationary curved plate , when jet strikes at one end (for symmetrical plate)

• Consider a fluid jet from a nozzle which strikes on a curved plate as shown. Let, • V = velocity of jet • d = dia of the jet • θ = angle made by the jet with horizontal before and after striking the plate • The fluid jet after striking the plate get deflected as shown and comes out with the

same velocity in the tangential direction of the curved plate. According to newtons law, force exerted is equal to the rate of change in momentum

• Fx = Rate of change in momentum in the horizontal direction = mass striking per sec x change in velocity = ρ a V [ V cosθ – (- V cos θ)] = ρ a V² [cosθ + cos θ] = 2 ρ a V² cosθ • Fy = Rate of change in momentum in the vertical direction = ρ a V [ V sinθ – V sin θ] = 0 • The resultant force exerted on the vane , F = • The direction of the resultant force is given by, • Work done = Force x Distance = ρ a V² x 0 =0 • Efficiency = Output of jet / Input of jet = Workdone per sec/ KE per sec = 0

22FyFx

Fx

Fy1tan


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Figure


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Force exerted by the jet on a stationary curved plate , when jet strikes at one end ( for unsymmetrical plate)

• Consider a fluid jet from a nozzle which strikes on a curved plate as shown. Let, • V = velocity of jet • d = dia of the jet • θ = angle made by the jet with horizontal before striking the plate • ϕ = angle made by the jet with horizontal after striking the plate • The fluid jet after striking the plate get deflected as shown and comes out with

the same velocity in the tangential direction of the curved plate. According to newtons law, force exerted is equal to the rate of change in momentum

• Fx = Rate of change in momentum in the horizontal direction = mass striking per sec x change in velocity = ρ a V [ V cosθ – (- V cosϕ)] = ρ a V² [cosθ + cosϕ] • Fy = Rate of change in momentum in the vertical direction = ρ a V [ V sinθ – V sinϕ] = ρ a V² [sinθ + sinϕ] • The resultant force exerted on the vane , F = • The direction of the resultant force is given by, • Work done = Force x Distance = ρ a V² x 0 =0 • Efficiency = Output of jet / Input of jet = Workdone per sec/ KE per sec = 0

22FyFx

Fx

Fy1tan


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N 13907.2 F

08.241.04

1000F gives, (1)eqn Therefore

42.08m/sV

1009.812

V0.95

2gH

V

V

V

Velocity lTheoretica

Velocity Actual C

...(1)........................................ V1.04

1000 F Therefore,

same) are nozzle andjet ofdiameter theassume ( m 0.1 d

1000

V a F : Ans

0.95. asgiven is velocity oft coefficien The plate. verticalfixed aon water ofjet by the exerted

force theFind 100m. is nozzle centre at the water of head theand 100mm is nozzle theof

diameter The fitted. is nozzle a which of end at the pipe a through flowing isWater :1 Prob

22

th

v

22

2


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N 2070.4

60sin250.0044171000

sinθsinθρaV

sinθFF

plate, the tonormaldirection in thejet by the exerted force (2)

N 2390.7

sin60250.0044171000

sinθρaVF

plate, the tonormaldirection in thejet by the exerted force (1)

60 θ

25m/s V

0.004417m0.0754

πd

4

π a

0.075md : Ans

jet. theofdirection in the (2)

plate the tonormaldirection in the (1) plate on thejet by the exerted force

theFind .60 is plate andjet ebetween th angle the way that asuch in plate

fixed a strikes 25m/s of velocity a with 75mm dia of water ofjet A :2 Prob

22

2

nx

2

2

n

0

222

0


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4711.15NF

cos60][1400.0019631000 F

]cosθ[1ρaV F

(-Vcos-ρaV[V F jet, ofdirection in the exerted Force

60120-180 θ-180 deflection of Angle

m/s 40V

0.001963m 0.054

π a Therefore

0.05m. d . Ans

plate. curved theofoutlet at the 120 of anglean

through deflected isjet theif jet, theofdirection in the water ofjet by the

exerted force theFind centre. at the plate lsymmetrica fixed curved a strikes

40m/s, of velocity a with moving 50mmdiameter of water ofjet A :3 Prob

x

2

x

2

x

x

0

22

0

θ)]


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N

V

N

V

m

13.628Fy

]20sin[sin3030004417.01000 Fy

]sin[sinaV Fy

)]sin([Vsin aV Fy

2.7178Fx

]20cos[cos3030004417.01000 Fx

]cos[cosaV Fx

)]cos([Vcos aV Fx

20

30

30m/s V ,Velocity

004417.00.0754

a Therefore

0.075m d Ans.

direction. verticaland horizontal in the plate on thejet by the

exerted force theFind .horizontal the to20 of anglean at plate theleavesjet The

.horizontal with the30 of anglean at end oneat lly tangentiaplate fixed curved a

strikes 30m/s, of velocity a with moving 75mm dia of water ofjet A :4 Prob

2

2

2

2

0

0

22

0

0


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Force exerted by the jet on a moving plate held normal to the jet

• Consider a fluid jet from a nozzle which strikes on a flat plate as shown. Let, • V = velocity of jet • u= velocity of plate • Relative velocity of jet with respect to plate = V-u • d = dia of the jet • The fluid jet after striking the plate get deflected to 90° as shown. Hence the component

of velocity in the direction of jet becomes zero after striking the plate. • According to newtons law, force exerted is equal to the rate of change in momentum • Force exerted by the jet on the direction of flow is given by Fx and perpendicular to the

direction of flow is given by Fy • Fx = Rate of change in momentum in the direction of flow = mass striking per sec x change in velocity = ρ a V-u [ V-u -0 ] = ρ a (V-u)2

• Fy = 0 ( Force exerted is opposite to each other. Hence cancels) • Work done = Force x Distance/Time = ρ a (V-u)² x u =0

2

2

2

x

VρaV2

1

uu-Vρa

mV2

1

uF

secper jet theofInput

secper jet by the Workdone

jet theofInput

jet theofOutput jet theof Efficiency


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Force exerted by the jet on a moving plate held inclined to the jet

• Consider a fluid jet from a nozzle which strikes on a flat plate as shown. Let, • V = velocity of jet • u= velocity of plate • Relative velocity of jet with respect to plate = V-u • d = dia of the jet • θ = angle between the jet and the plate • Fn = force exerted by the jet on the plate in the direction normal to the plate • Fn = Rate of change in momentum in the direction normal to the plate = mass striking per sec x change in velocity = ρ a (V-u) [(V-u) sin θ -0 ] = ρ a (V-u)² sin θ • Fn can be resolved in to two components , Fx which is parallel to the direction of jet and Fy

perpendicular to the direction of jet • Fx = Fn sin θ = ρ a (V-u)² sin θ * sin θ = ρ a (V-u)² sin² θ • Fy = Fn cos θ = ρ a (V-u)² sin θ cos θ • Work done = Force x Distance = ρ a (V-u)² sin² θ x u

•

2

22

2

x

VρaV2

1

usinu-Vρa

mV2

1

uF



jet theofInput



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Force exerted by the jet on a moving curved plate, when jet strikes at the centre

• Consider a fluid jet from a nozzle which strikes on a curved plate as shown. Let, • V = velocity of jet • u= velocity of plate • Relative velocity of jet with respect to plate = V-u • d = dia of the jet • θ = angle made by the jet with horizontal after striking the plate

• The fluid jet after striking the plate get deflected as shown and comes out with the same velocity in the tangential direction of the curved plate. According to newtons law, force exerted is equal to the rate of change in momentum

• Fx = Rate of change in momentum in the direction of flow = mass striking per sec x change in velocity = ρ a (V-u) [(V-u) – (- (V-u) cosθ)] = ρ a (V-u) [(V-u) +(V-u) cosθ] = ρ a (V-u)2 [1 + cosθ] • Fy = Rate of change in momentum perpendicular to the flow = 0 ( Force exerted is opposite to each other. Hence cancels) • Work done = Force x Distance = ρ a (V-u)2 [1 + cosθ] x u • Angle of deflection of the plate = 180- θ

2

2

2

x

VρaV2

1

uCos1u-Vρa

mV2

1

uF



jet theofInput



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%8.28288.0

1515007854.010002

1

617.636

ρaV2

1

u Fjet theof Efficiency

secper jet theofenergy Kinetic

secper jet by the doneWork


secper jet theofOutput jet theof Efficiency

KW 3.81 W3817.02 secper jet by the done Work jet theofPower

Nm/s 3817.02617.636u F secper jet by the done Work

N 17.6366)(15007854.00001F

u)ρa(VF

6m/su ; 15m/sV

007854.0d4

a ; 0.1m d : Ans

jet. theof efficiency (4)jet theofpower (3) secper plate on thejet by the donework

The (2) plate on thejet by the exerted force The (1) Find jet. thefromaway and

jet theofdirection in the 6m/s of velocity a with moving is plate The 15m/s. of

velocity aith normally w plateflat a strikes 10cm dia of water ofjet A :5 Prob

22

x

x

2

x

2

x

22

V

m


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%5.12125.0

3030d4

π1000

2

1

5.7453

VρaV2

1

5.7453

jet theofenergy Kinetic

poweror secper workdone

Input

Outputjet theof efficiency (6)

watt5.7453secper plateon jet by the workdonejet theofpower (5)

Nm/s5.74531591.496plate/sec

by moved Distance jet ofdirection in the Force sec,per plateon jet by the workdone(4)

496.91N45cos74.702cosFF jet, ofdirection thelar toperpendicu force the(3)

1N9.49645sin74.702sinFF jet, ofdirection on the force the(2)

74.70245sin1530d4

π1000

sinuVρaF jet, the tonormal Force (1) :Ans

jet theof efficiency (6)

jet theofpower (5)

secper plateon jet by the workdone(4)

jet ofdirection thelar toperpendicu force the(3)

jet ofdirection on the force the(2)

plate the tonormal force the(1)

Find 15m/s. of velocity a with moving is plate The jet. theof axis the to45at inclined is

whichof normal theplate,flat a strikes 30m/s of velocity a havingjet dia 7.5cmA :6 Prob

22

2

nx

nx

22

2

n

0

N


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%4.56564.0

2020004417.010002

1

838.1250

2

1

F

2

1

F

secper jet ofenergy Kinetic

secper vaneon thejet by the workdone

Input

Outputjet theof efficiency (3)

04.10003w04.10003/04.10003838.1250F

secper vaneon thejet by the workdonejet theofpower (2)

38.1250F

15cos1)820(004417.01000F

cos1)(F

cos)()()(F

F jet, ofdirection in the vaneon the exerted force the(1)

15180165 Therefore ;180165

8m/su 20m/s;V

m004417.0d4

a ; 0.075md : Ans

jet theof efficiency (3)jet theofpower (2)jet ofdirection the

in vaneon the exerted force the(1) findsmooth be to vane theAssume .165 of anglean through

deflected isjet The jet. ofdirection in the 8m/s of velocity a with moving is vanecurved The

20m/s. of velocity a with centre itsat vanecurved a strikes cm 7.5 dia of water ofjet A :7 Prob

22

x

2

x

x

x

02

x

2

x

x

x

00000

22

0

VaV

u

mV

u

KWsNmu

N

uVa

uVuVuVa


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axis x with thedirection clockwisein is angle themeans

51237.1tan Therefore

237.114

32.17

F

F tanaxiswith x resultant by the made Angle

27.2232.1714FFF on vane, forceResultant

downwards) acting is F meanssign (-ve 32.1725sin60-08.0F

yin velocit Change secper stricking Mass F

1425cos60-308.0F

yin velocit Change secper stricking Mass F

F jet, ofdirection in the Force

60

0.8kg/s secper flowing Mass

m/s52V 30m/s;V :Ans

.stationary is vane theif vane,on the force

resultant theofdirection and magnitude thecalculate 0.8kg/s. is nozzle thefrom

discharge theIf 25m/s. ofity mean veloc a with leaves and 30m/s ofvelocity

a shock witht lly withou tangentiaentersit which plate curved aby direction

original its from 60 through deflected is nozzle a from water ofjet A :8 Prob

01

x

y

222

y

2

x

y

0

y

y

0

x

x

x

0

21

0

ve

N

N

N


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N

g

829.581.9

27.18930 tunit weighper F

81.9 secper strickingjet of Mass

5))]220)cos-(-(50-20)-[(50 secper strickingjet of Masstunit weighper F

secper strickingjet of Mass

))]u)cos-(-(V-u)-[(V secper strickingjet of Masstunit weighper F

jet theoft Unit weigh

yin velocit Change secper strickingjet of Masstunit weighper F

equation. below thefromout found becan

t unit weighper Fx But .determined becannot secper strickingjet of mass the

as found becannot Fx the thusgiven.not isjet theofdiameter problem In this

yin velocit Change secper strickingjet of MassF

25 ; 0 ; 20m/s u ; 50m/sV : Ans

flow. theoft unit weighper developedpower and workdone theCalculate

direction. and magnitudein forceresultant thedetermine Also it. across and

velocity vane theofdirection in the acting forces ofcomponent theCalculate

50m/s. isjet of velocity The .25 is angleoutlet theand degree zero is angleinlet

The jet. ofdirection in the 20m/s of velocity a with moving is vaneA : 9 Prob

x

x

x

x

x

00

0


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watts116.58 in watts secper Workdone developedPower

116.58Nm/s205.829

tunit weighper F per weight secper Workdone

30.122217.0tan

2217.0829.5

292.1

F

F tan axis with x forceresult ofDirection

917.5292.1829.5FFForceResultant

292.181.9

30sin25)-[0 tunit weighper F

81.9 secper strickingjet of Mass

u)sin25)-(V-[0 secper strickingjet of Masstunit weighper F

jet theoft Unit weigh

yin velocit Change secper strickingjet of Masstunit weighper F

x

01

x

y

222

y

2

x

y

y

y

u

N

N


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Practice Problem

Prob 10: A jet of water of diameter 50mm moving with a velocity of 25m/s impinges on a fixed curved plate tangentially at one end at an angle of 30° to the horizontal. Calculate the resultant force of the jet on the plate if the jet is deflected through an angle of 50°. Take g as 10m/s2

Ans : Fx=849.7 N; Fy= -594.9 N; FR=1037N; Resultant angle = 35° with horizontal


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Force exerted on an unsymmetrical moving curved plate when jet strikes tangentially


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• If the jet strikes tangentially, the loss of energy due to the impact of the jet will be zero. – Let V1 and V2 = Velocity of the jet at inlet and outlet respectively – u1 = u2 = u . Velocity of the plate – Vr1 and Vr2 = Relative velocity of jet and plate at inlet and outlet – α = angle made by V1 with the direction of motion of plate. Also known as guide

blade angle – β = angle made by V2 with the direction of motion of plate – θ = angle made by Vr1 with the direction of motion of plate. Also known as vane

angle at inlet – ϕ = angle made by Vr2 with the direction of motion of plate. Also known as vane

angle at outlet – Vw1 and Vf1 = Components of V1 in the direction of motion and perpendicular to

the direction of motion. Vw1 is known as whirl at inlet. Vf1 is known as flow at inlet

– Vw2 and Vf2 = Components of V2 in the direction of motion and perpendicular to the direction of motion. Vw2 is known as whirl at outlet. Vf2 is known as flow at outlet

– The triangles ABD and EGH are known as inlet and outlet velocity triangles respectively 18-May-17 31 NISHANT, ISSAT

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• Construction of velocity triangle at inlet 1. Take point A and draw a line AB = V1 in magnitude , making an angle α with the

horizontal line AD as direction

2. Draw a line AC=u1 in magnitude. Join C to B. CB represents the relative velocity

3. From B draw a vertical line BD in the downward direction to meet the horizontal AC produced at D. BD represents the velocity of flow and CD represents the velocity of whirl

If the vane is assumed to be smooth or water enters and leaves without shock then Vr1=Vr2

For equal vane velocity in inlet and outlet , u1 = u2 = u

• Construction of velocity triangle at outlet 1. Draw EG in the tangential direction such that EG=Vr2.

2. From G draw GF equals to U2 in magnitude. Join EF which represents the velocity of vane at outlet in magnitude and direction.

3. From E draw a vertical line EH to meet the line GF produced at H. EH represents the velocity of flow at oulet and FH represents the velocity of whirl at outlet


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2

wwr

ww

r

wwrwwr

x

ww

r

wwrwwr

x

wwr

wwrx

wwrxx

wwrx

wrx

2

21wwrx

2w21w1r1x

2w21w1r1x

rrrx

211x

x

11

211

21

1

211211

21

1

211211

211

211

211

211

11

211

211

VaV2

1

u VV aV

jet theofenergy kinetic Initial

on vane secper workdone

Input

Output jet, theof Efficiency 6

u VVaV

u VV aV

secper Mass

u VV aV

secper striking fluid of mass

u F vaneon the massunit per secper Workdone5

g

u VV

g aV

u VV aV

g secper Mass

u VV aV

secper striking fluid ofweight

u F vaneon thet unit weighper secper Workdone4

u VV aV vaneon the secper WorkdonePower 3

u VV aV u F vaneon the secper Workdone2

VV aV F , as written becan Fgenerally Thus

VV aV Fbecomes, (1)eqn Then , obtuse is If

V aVF becomes, (1)eqn Then

zero. becomes Vw ,90 If angle. acute is only when satisfies (1)eqn

uu Since .....(1).......... VV aV F

uVu-V aV F

uVu-V aV F

cosVcos V aV F

cosuVcos u-V u)-a(V F

yin velocit Change secper striking water of Mass F 1


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Force exerted on an unsymmetrical moving curved plate when jet strikes tangentially


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r2f1

r1

1w1

1f1

1w1

f1

r2r1

211

V 11.14m/s87.37sin

84.6

sin

VV ABC, From

87.37

10-79.18

84.6 tanTherefore

s/m79.1820osc20cos VV

s/m84.620sin20sin VV where

u-V

V

CD

BD tanABD,In

shock) without leaves and enters water (since VV

60130-180 ;20 10m/s;uu 20m/s;V : Ans

secper vane thestricking water oft unit weighper secper workdone(2)

shock without vane theleaves and enters water that theso angles, vane(1)

Calculate outlet.at

vaneofmotion ofdirection the to130 of anglean at leaves andinlet at vaneofmotion

ofdirection with the20 of anglean makesjet The m/s. 10 of velocity a with moving

is which vane,curved a strikes m/s 20 of velocity a having water ofjet A :10 Prob


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NNm

u

u

/24.20

067.1cosVV,79.18V since 109.81

06.179.81

g

VV wateroft unit weighper secper Workdone)2(

56.6 Therefore,

-05sin

10

05-180sin

11.14

-sin

u2

-180sin

V EFG, From

2r2w2w1

w2w1

r2


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12060-180GEF

/10uGF

/10VEG angle,outlet tri From

/10VV

/20VV

/10uVAC-ABV ngle,inlet tria From

shock without vane theleaves and enters water Assuming VV

vane)ofmotion ofdirection with the60 with leavesjet since ( 6060180

direction) samein move vaneandjet since ( 0

/10uu ;/20V

001963.0d4

a ; 0.05md :Ans

jet by the secper workdone(2)

vaneofdirection in the vaneon thejet by the exerted force The (1)

:Determine outlet.

at vaneofmotion ofdirection the to60 of anglean at vane theleavesjet vane.Theof that as

direction samein moves 20m/s of velocity a having , 50mmdiameter of water ofjet A :11 Prob

2

r2

r2r1

1w1

11r1

r2r1

211

22

sm

sm

sm

sm

sm

smsm

m


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Nm/s 2944.510 294.45 u Fx secper Workdone2

45.29450201001963.01000

angle obtusean is here VVaVFx

Fx motion, ofdirection in thejet by the exerted Force 1

/560cos1010cosVuGH-GFV

60 Therefore ,-120sin

10

sin60

10

-120sin

GF

sin60

EG rule,sin From GEFIn

w2w1r1

r22w2

N

sm


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18.36 solving ;Vr2

u2cos BCDIn

Vr2Vr1

53.79 solving ;u1-vw1

Vf1 tanBCDIn :Hint

shock. without vane theleaves and enters water that thesooutlet andinlet at angles

vane thedetermine andoutlet andinlet at triangle velocity theDraw outlet.at vaneof

motion ofdirection the to90 of anglean at leaves andinlet at vane theofmotion of

direction with the30 of anglean makesjet The 20m/s. of velocity a with moving

is which vane,curved a strikes 40m/s of velocity a having water ofjet A :12 Prob


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m/s 11.41cos17.9812cosVV

7.41m/sV solving

98.17sin

V

03-180sin

V

sin

BC

-180sin

AB ABC,In

17.98 vane,ofmotion ofdirection theinlet withat jet of Angle

17.98 Solving,

-30sin

u

03-180sin

V

-30sin

AC

-180sin

AB ABC,In

lsymmetrica is vane thesince 30

60120-180 Therefore

120deflection of Angle

5m/suu 12m/s;V : Ans

l.symmetrica andsmooth be to vane theAssume

striking. water oft unit weighper secper workdone theFind magnitude. anddirection

in exit at jet theof velocity absolute theis What inlet.at shock no is e that thersojet

theof angle thefind , 5m/s with moving is vane theIf . 120through

jet edeflect th toshaped on vane impinges 12m/sat moving water ofjet A :14 Prob

1w1

r1

r11

11

211


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Nm/s 537.65

81.9

417.141.11VV

aV

VVaV tunit weighper secper Workdone

m/s96.3VVV outlet,at jet of Magnitude

55.11007.69-180 vaneofmotion

ofdirection h theoutlet witat Vby made angle

07.69 Therefore,

V

Vtan

3.705m/ssin307.41sin30 VV

m/s 1.417 5-03cos7.41V

u-cos VV EFH,In

7.41m/sVV

-180 vaneofmotion ofdirection h theoutlet witat Vby made angle The

Vby given isoutlet at jet of velocity absolute The

w2w1

r1

w2w1r1

2

w2

2

f22

2

w2

f2

r2f2

w2

2r2w2

r2r1

2

2

ug

ug


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50.5% Efficiency ;KW 6.703 Power

1340.6 F 2.07m/s;V 15m/s;V ,45 :Hint

vane theof Efficiency 3

and vane,on the exertedpower 2

motion ofdirection in the vaneon thejet by the exerted Force 1

Find smooth. be to vane theAssuming 100mm. isjet of diametre The .135 through deflected is

jet that theshaped so is vaneThe inlet.at jet theof that asdirection same in the m/s 5 of velocity a

withmoving is which vanecurved a strikes m/s 15 of velocity a having water ofjet A :15 Prob

xw2w1


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Force exerted by the jet on a hinged plate

• Consider jet striking on a vertical plate at centre, hinged at O

• The plate will swing through some angle due to the force exerted by the plate – X = distance of centre of jet from hinge O

– θ = angle with which the plate swings

– W = weight of the plate. Acting through C.G ( point A’)

• Dotted lines shows the position of plates before jet strikes

• The point A changes to A’ after the jet strikes. Distance OA=OA’= x

• After the jet strikes the plate, for the plate to be in equilibrium , the moment of all forces acting about the hinge must be zero


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swing of angle therepresents where

W

aVsin isThat

sin x W x aV

0sin x W - x aV

0 M mequilibriuin be toplate For the

sin x W sin OA' W weight todue hingeabout Moment

A'about actingbody theof Weight 2.

x aVosc

OAoscaVOBoscaV

OBCosaVOBFF force todue hingeabout Moment

CosaV F , plate the tonormal acting Force 1.

strike.jet after the plate on the acting forces twoare There

2

2

2

222

2

nn

2

n


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N5.2420

1049 P Therefore

20 P 10 V a

20 P 10 F

0 20 P - 10 F , hinge about the moments the takingTherefore

m,equilibriuin plate thekeep order toin Ppoint through applied be force alet (1)

10m/sV ; 98.1N W ; 0.00049ma ; m .025 d : Ans

? water.ofjet by exerted force the todue calwith verti

plate theofn inclinatio thebe l what wilfreely,deflect toallowed is plate

theIf cal.kept verti is plate that theso plate theof edgelower at the applied

bemust force What plate. the tonormal strikesjet The N. 98.1 is plate the

of weight The edge. horizontal topitson hinge aby y verticallsuspended is

whichplate theof centre at the m/s 10 of velocity a with thicknessuniform

of plate square 20cm x 20cm a strikes 25mm dia of water ofjet A :16 Prob

2

2


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30 Therefore

sin101.98101000049.01000

sin W cos

OA cosV a

GD W OBFn

0 GD W OBFn

gives, hinge about the forces 2 theseof moments Taking

plate theof weight W todue

force (b) andFn todue force (a) are mequilibriuat plate on the acting forces two

plate, by the made swing thebe let )2(

2

2

OG


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Force exerted on a series of vanes

• Force exerted by the jet on a single vane is not practically feasible inorder to run a turbine

• In actual practice a large number of vanes are mounted on the circumference of the wheel at a fixed distance apart

• The force exerted by the jet will be enough for the first vane to move , and the second vane mounted on the wheel appears before the jet. The jet again exerted force on the second vane. The process will continue and the turbine will rotate continuously at constant speed.


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• For a single vane the mass of water stricking per sec is given by ρa(V-u). But for a series of vanes the mass of water stricking per sec is given by ρaV. This is because for a series of vanes the jet from nozzle is always in contact with the vanes as a number of vanes is considered.

Mass of water stricking per sec = ρaV

Velocity with which jet strikes the plate = V-u

Fx= Mass per sec x [initial velocity – final velocity]

Fx = ρaV x [ (V-u) – 0 ]

Fx = ρaV x (V-u)

Work done by the jet on the series of vanes = Fx x u

= ρaV [(V-u)] x u


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50%or 2

1

2u

1-2u

2u

u-2u

2u

u]-u[2u2

as, efficiencymax gives 1eqnin 2u V of value thengsubstituti

jet of velocity of half is vaneofcity when velooccurs efficiencymax Thus

2

Vuor

2

4uV

0V

4u-V2

0V

2u-uV2

du

d

0V

u]-u[V2

du

d

0du

d jet,given a of efficiencyMax For

1.................................................V

u]-u[V2

VaV2

1

u u]-aV[V

mV2

1

u F

secper energy Kinetic

secper doneWork ,Efficiency

2

2

2

2

2

22

2

x


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60120-180 ;30 ; m/s 20uu ; m/s 35V : Ans

efficiency the3)

vanes theentering water oft unit weighper workdone the2)

shock without leaves and entersr that wateso tips vaneof angle the1)

:find also andoutlet andinlet at rianglesvelocity t

theDraw .120 of anglean at leavesjet theand entering when vanesofmotion of

direction the to30 of anglean makesjet The 20m/s. of velocity a with moving

vanesof series aon impinges 35m/s of velocity a having water ofjet A : Prob

211

0


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%74.99

g2

V

4.62

mg

mV2

1

4.62

tunit weighper energy Kinetic

tunit weighper doneWork ,Efficiency )3

s/Nm4.6281.9

203.031.30

g

uVVWorkdone

gaV

uVVaV entering water oft unit weighper Workdone

3.0tan

VV Therefore .

V

Vtan

524.025.20.251sin Vsin V )2

1.25 ,outlet at angle vaneSolving

60Sin

20

120Sin

25.20

Sin

V

60Sin

u

120Sin

V

rulesin by ngle,ocity triaoutlet vel theFrom

25.2060sin

5.17

sin

1VfVV

60 inlet,at angle vaneTherefore

.2031.30

5.17

uV

V tan

s/m31.3030cos35V

s/m5.1730sin35V

gle,city trianinlet velo theFrom 1)

2

12

1

2w1w

1

2w1w1

2f2w

2w

2f

2r2f

222r

2r1r

11w

1f

1w

1f


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Force exerted on a series of Radial curved vanes

2w21w11

1w11

2w21w11

2w21w11

2w21w11

2w21w11

2w21w11

2w211w11

2w21221

1w11111

1111

2

1

u Vu VV a secper done work generalIn

90 is when applicable iseqn This .u VV a Workdone

obtuse is when applicable iseqn This .u Vu VV a Workdone

acute is when applicable iseqn This .u Vu VV a Workdone

R VR VV a Workdone

R VR VV a Workdone

T locity Angular ve Torque secper on wheel Workdone

R VR VV a T

R VV a -R VV a T , Therefore

R VV a - R cosVV a

vaneofoutlet from Radius outlet at Momentum secper momentum Final

R VV a R cosVV a

vaneofinlet from Radius inlet at Momentum secper momentum Initial

secper momentum Final - secper momentum Initial T

momentumangular of change of Rate T wheel,on the water by the exerted Torque

r vsince R u , R u , Therefore

rotates wheelh the with whicspeedAngular

vane theofoutlet from wheelof Radius R

vane theofinlet from wheelof Radius R Let,

shown as wheelaon mounted vanescurvedradially of series aConsider

different. isoutlet andinlet at vaneof radius

theas equalnot isoutlet andinlet at vaneradial afor velocity vaneThe


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max be toV beingfor minimum be should or minimum be should )3

maximum be should V and V toadded is V when 2)

minimum is V when 1)

max when, is that concluded becan it equations above theFrom

V

V1

V

V-V

VaV2

1

V-VaV2

1

secper KE Initial

secper doneWork be,can efficiency Hence

V-VaV2

1

V-Vm2

1mV

2

1mV

2

1 secper doneWork

secper KE Final - secper KE Initial secper doneWork

as, written be alsocan secper workdoneTherefore sec.per

jet ofenergy kineticin change the toequal be willsecper workdone

the vanes,over the flowing isn water energy whe of loss no is thereIf

V

u Vu V2, Efficiency

VaV2

1

u Vu VaV

secper energy Kinetic

secper doneWork , Efficiency

w2

w2w1w2

2

2

1

2

2

2

1

2

2

2

1

2

11

2

2

2

11

2

2

2

11

2

2

2

1

2

2

2

1

2

1

2w21w1

2

11

2w21w11


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s/m 235.594.2025.0Ru

s/m 47.1094.205.0Ru Therefore

m25.0R ;m5.0R

;50130180 ;s/m5V ;20

s/rad94.2060

2002 rpm; 200 N ;s/m30V : Ans

wheel theof Efficiency 3)

wateroft unit weighper done Work 2)

outlet andinlet at angles Vane 1) :Determine

ly.respective 0.25m and 0.5m are wheel theof

radiiinner andouter The direction. radial ain outward from flowing isWater

wheel. theofoutlet theo tangent t the to130 of anglean at 5m/s of velocity a

with wheel theleaves andinlet at wheel theo tangent t with the20 of anglean

makesjet The 200rpm.at rotating is which wheelaon mounted vanescurved

radially of series a strikes 30m/s of velocity a having water ofjet A :Prob

22

11

21

2

1


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%32.69

032

1

235.5214.310.4728.19

V2

1

uVuV

VaV2

1

uVuVaV

Energy Kinetic Initial

secper doneWork ,Efficiency 3

N/Nm 8.3181.9

235.5214.310.4728.19

g

uVuV

gaV

uVuVaV

sec,per striking water oft unit weighper Workdone2

385.24 outlet,at angle vaneTherefore

435.0214.3235.5

3.83

uV

V

HF

EH tanEFH,In

m/s 83.350sin5sinVV

m/s 214.350cos5cosVV ,GEH From

07.30 inlet,at angle vaneTherefore

579.010.47-19.28

10.26

u-V

V

CD

BD tanCBD,In

m/s 26.1020sin03sinVV

m/s 19.2820cos03cosVV ABD, From 1

22

1

2w21w1

2

11

2w21w11

2w21w1

1

2w21w11

2w2

f2

2f2

2w2

1w1

f1

1f1

1w1


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Euler’s equation

dS

dZ g

2

dAA

cosmg motion ofdirection in the weight ofcomponent 3

dAdPP- dPP pressure todue Force 2

dAP P pressure todue Force 1

aremotion ofdirection in the acting forces The

elevation dZ Z velocity, dVV intensity, pressure dPP area, dA A

CD, ofsection at theLet

elevation Z velocity, V intensity, pressure P area, A

AB, ofsection at theLet

shown as streamtube ofsection a Consider


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Euler’s equation

equation sEuler' isequation above eTh

0dS

dZg

dS

dP1

dS

dVV

dS

dZg

dS

dP1

dS

dVV

becomes,equation theAdSby

dZA g - AdP- dVAV

becomes,equation thegsimplifyin

dS

dZ g

2

dAA dAdPP-dAPdVAV

dS

dZ g

2

dAA dAdPP-dAP momentum of change of Rate

momentum of change of rate toequal is force law, newtons toAccording

dS

dZ g

2

dAA dAdPP-dAP

by,given ismotion ofdirection on the acting forceresultant Therefore


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Classification of Hydraulic Machines


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Introduction to turbines

• Turbines is defined as the hydraulic machines which convert hydraulic energy into mechanical energy.

• This mechanical energy is used to rotate an electric generator which is directly coupled to the turbine. Electric generator converts the mechanical energy into electrical energy.

• The energy obtained from turbines is also known as hydroelectric power


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General layout of Hydroelectric plant

Hydro-electric power plant which consists of :

1. A dam constructed across a river to store water.

2. Pipes of large diameters called penstocks, which carry water under pressure from the storage reservoir to the turbines. Penstock is made of steel or reinforced concrete

3. Turbines having different types of vanes fitted to the wheels.

4. Tail race, which is a channel which carries water away from the turbines after the water has worked on the turbines. The surface of water in the tail race is also known as tail race


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Heads of a turbine

Two different heads are available

1. Gross head : It is the difference between head race level and tail race level. Denoted by Hg

2. Net head: It is the actual head available at the inlet of the turbine after considering all the losses.

Net head = Hg - Hf

Where Hf is the head loss due to friction which is given by

gD2

LV f4H

2

f


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Efficiencies of a Turbine The followings are the important efficiencies of a turbine.

(a) Hydraulic Efficiency (ηh)

(b) Mechanical Efficiency (ηm)

(c) Volumetric Efficiency (ηv)

(d) Overall Efficiency (ηo)

Hydraulic Efficiency (ηh)

• It is defined as the ratio of power given by water to the runner of a turbine to the power supplied by the water at the inlet of the turbine.

turbineat the available headNet H

g V a vanes thestriking water of weight W Where

KW 1000

HWor KW

1000

VV a 2

1

WP

by,given ish power whic water theis WP

KW 1000

uV uVV a by given is turbineflow radialfor RP

KW 1000

uV VV a by given is binepelton turfor RP

Power Runner theis RP Where

2

2w21w1

w2w1


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• The power at the inlet of the turbine is more and this power goes on decreasing as the water flows over the vanes of the turbine due to hydraulic losses as the vanes are not smooth. Hence the power delivered to the runner of the turbine will be less than the power available at the inlet of the turbine.

• Mechanical Efficiency (ηm)

The power delivered by water to the runner of a turbine is transmitted to the shaft of the turbine. Due to mechanical losses, the power available at the shaft of the turbine is less than the power available at the runner of the turbine.

The ratio of the power available at the shaft of the turbine to the power delivered to the runner is defined as mechanical efficiency.


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3. Volumetric Efficiency (ηv)

The volume of the water striking the runner of a turbine is slightly less than the volume of the water supplied to the turbine. Some of the volume of the water is discharged to the tail race without striking the runner of the turbine.

The ratio of the volume of the water actually striking the runner to the volume of water supplied to the turbine is defined as volumetric efficiency.

4. Overall Efficiency (ηo)

It is defined as the ratio of power available at the shaft of the turbine to the power supplied by the water at the inlet of the turbine.


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Classification of Hydraulic Turbines

1. According to the type of energy at inlet: (a) Impulse turbine, (b) Reaction turbine • If at the inlet of the turbine, the energy available is only kinetic energy,

the turbine is known as impulse turbine. The pressure is atmospheric from inlet to outlet of the turbine.

• If at the inlet of the turbine, the water possesses kinetic energy as well as pressure energy, the turbine is known as reaction turbine. As the water flows through the runner, the water is under pressure and the pressure energy goes on changing into kinetic energy.

2. According to the direction of flow through runner: (a) Tangential flow turbine (b) Radial flow turbine, (c) Axial flow turbine, (d) mixed flow turbine • If the water flows along the tangent of the runner, the turbine is known at

tangential flow turbine. • If the water flows in the radial direction through the runner, the turbine is

called radial flow turbine. • If the water flows from outwards to inwards, radially, the turbine is known as

inward radial flow turbine, on the other hand, if water flows radially from inwards to outwards, the turbine is known as outward radial flow turbine.


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• If the water flow through the runner along the direction parallel to the axis of rotation of the runner, the turbine is called axial flow turbine.

• If the water flows through the runner in the radial direction but leaves in the direction parallel to the axis of rotation of the runner, the turbine is called mixed flow turbine.

3. According to the head at the inlet of turbine:

(a) High head turbine, (b) Medium head turbine, and

(c) Low head turbine

4. According to the specific speed of the turbine:

(a) Low specific speed turbine (b) Medium specific speed turbine,

(c) High specific speed turbine


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Pelton wheel ( Pelton turbine) • It is an tangential flow impulse turbine

• The jet flows tangentially over the vanes

• The energy available at the inlet of the turbine is KE only

• Pelton turbine is a high head turbine

• The water from the reservoir flows through the penstocks at the outlet of which a nozzle is fitted. The nozzle increases the kinetic energy of the water flowing through the penstock

• The main parts of a pelton turbine are

1. Nozzle and flow regulating arrangement ( spear)

2. Runner and buckets

3. Casing

4. Breaking Jet

5. Tail race


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Pelton Turbine


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1. Nozzle and flow regulating arrangement ( spear)

• The amount of water striking the buckets (vanes) of the runner is controlled by providing a spear in the nozzle.

• The spear is a conical needle which is operated either by a hand wheel or automatically in an axial direction depending upon the size of the unit.

• When the spear is pushed forward into the nozzle the amount of water striking the runner is reduced. On the other hand, if the spear is pushed back, the amount of water striking the runner increases.


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Runner and Buckets

• consists of a circular disc on the periphery of which a number of buckets evenly spaced are fixed. The shape of the buckets is of a double hemispherical cup or bowl. Each

bucket is divided into two symmetrical parts by a dividing wall which is known as splitter. • After striking the splitter the jet divides in to two equal parts and the jet comes out through the outer edge of the bucket • The jet gets deflected through an angle of 160° to 170° after striking the vane • Buckets are made of caste iron, cast steel, bronze or stainless steel


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Casing

• Turbine is enclosed in a casing

• Casing is provided to prevent the splashing of the water and to discharge water to tail race.

• It also acts as a safeguard against accidents.

• It is made of cast iron or fabricated steel plates.


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Breaking Jet

• When the flow of jet to the bucket is stopped , the runner still runs for a long period due to inertia force acting.

• To stop the runner in a quick time, a small nozzle is provided which directs the jet of water on the back of the vanes. This jet of water is called breaking jet.

Tail race

• The water after striking the buckets of is collected in the tail race


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Velocity triangles , Workdone and Efficiencies of a Pelton wheel

• The jet of water strikes the splitter tangentially and splits into two equal parts.

• The jet flows over the inner surfaces and comes out through the outer edge.

• The splitter is the inlet part and the outer edge is the outlet part of the bucket.

• The inlet velocity triangle is drawn at the splitter and outlet velocity triangle is drawn at the outer edge of the bucket


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g

u V V

gV a

u V V V a sect unit weighper jet by the Workdone8

u V V V a

u Fx secper jet by the Workdone7

) of 90for ( V V a Fx

) of angle obstusefor ( V V V a Fx

) of angle acutefor ( V V V a Fx

vane,ofmotion ofdirection in thejet by the exerted Force 6

ucosVV

VV

outlet,at triangle velocity For the 5

0 0,

u-V V

VV

where,line a be inlet willat triangle velocity The 4

60

DNuu vane,ofVelocity 3

0.99or 0.98 C e wher

2gHC becomescity inlet velo lTheoretica given, is C If

2gH city inlet velo lTheoretica 2

D 2g

LV 4f loss Head H

head GrossH where

H - H H el,pelton whe on the acting headNet 1

w2w1

1

w2w11

w2w11

w11

w2w11

w2w11

2r2w2

r2r1

11r1

1w1

21

V

VV

P

2

1f

g

fg


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2

Vu,Therefore

0u2V

0u-uVdu

d

V

cos 12

0V

u cos 1 u-V2

du

d

0du

d,efficiencymaxFor

2............V

u cos 1 u-V2

V

u cosu-V u -V 2

V

u u-cosu-V V 2

gives, 1eqn in valuesabove thengSubstituti

u -cosu-Vu-cosVu-cosVV

VV where

1...........V

u V V 2

VV a 2

1

u V V V a

efficiency maximumfor Derivation 10

VV a 2

1

u V V V a

secper jet of KE

secper Workdone

secper jet of KE

secper Workdone , efficiency Hydraulic 9

1

1

2

12

1

2

1

1

h

2

1

1

2

1

11h

2

1

11h

1r12r2w2

1w1

2

1

w2w1

2

11

w2w11h

2

11

w2w11h

h


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deflection of angle -180

givennot if 165 be todeflection of angle Take 12

rpmin speed wheel- N and wheelof diametrepitch - D e wher

60

DNuby found be alsocan or wheel vaneofVelocity

givennot if 0.45 Take

0.48 to0.43 from varies ratio. speed as known is re whe

2gH

u velocity.al theoritic theto velocity vaneof ratio theis ratio Speed 11

2

cos 1

V

2

V cos 1

2

V2

V

2

V cos 1

2

V-V2

Max Therefore

2eqn in2

V asu of value thengsubstitutiby obtained becan Max

jet of velocity of half is vaneofcity when velomaximum

is elpelton whe of efficiency Hydraulic that thefoundbeen hasit Thus

2

1

11

2

1

111

h

1h


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jet single a through hargeDisc

discharge Total jets ofNumber

jet oneby

given discharge the todischarge totalby thegiven isIt jets. ofNumber 15

m5.0152d

D15 Zrunner, ain buckets ofNumber 14

12 mNormally 14. to10 from ranges m of valueThe .d

D m by,given isIt

.djet theofdiameter the to

Delpelton whe ofdiameter pitch of ratio theas defined isIt . m ratioJet 13


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Numerical Problems in pelton turbine

Refer page no 862 to 877 ( RK. Bensal)


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Design of a Pelton turbine

Design of a pelton turbine means the following data is to be determined

1. Diameter of the jet (d)

2. Diameter of the wheel (D)

3. Width of the buckets = 5 x d

4. Depth of the buckets = 1.2 x d

5. Number of buckets on the wheel ,

18-May-17 NISHANT, ISSAT 81

d2

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24 buckets ofNumber

0.102m buckets ofDepth

0.425m buckets of width

m85.0d

m44.1D;60

DNu

62.33gH2CV

:Answer

0.98. toequal isvelocity

oft coefficien and 0.85 is efficiency Overall jet. theof velocity the times0.45

is buckets of velocity The power.shaft KW 95.6475 develops elpelton whe The

200rpm.at running when 60m of head afor designed be tois elpelton wheA : Prob

v1

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Pelton turbine bucket design


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Numerical questions in Design of pelton turbines

Refer following problems of R.K.Bensal

1. 18.11 (pg no 874)

2. 18.13 (pg no 875)


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Francis Turbine

Refer page no 877 to 903 (RK.Bensal)


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Radial Flow Reaction Turbines

• In radial flow turbines water flows in the radial direction

• Water may enter from outwards to inwards (inward radial flow turbine) or from inwards to outwards (outward radial flow turbine)

• In reaction turbines the inlet water possess both Kinetic and Pressure energy

• As the water flows over the runner a part of the pressure energy changes into pressure energy


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Parts of Radial flow turbine

1. Casing

2. Guide wheel

3. Runner

4. Draft tube


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1. Casing – Casing is spiral shape in which cross sectional area goes on decreasing

gradually

– In reaction turbine casing is always full of water

– Casing completely covers the runner

– Casing is made of concrete, caste steel or plate steel

2. Guide mechanism – It consist of a stationary circular wheel mounted with vanes known as guide

vanes

– The guide vanes allow the water to strike the vanes fixed on the runner without shock

– The width of the guide vane can be adjusted so that the amount of water striking the runner can be varied


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3. Runner – It is a circular wheel on which a series of radial curved vanes are fixed

– The surface of the vanes are made very smooth so that there is no loss of energy

– The radial curved vanes are shaped that the water enters and leaves without shock

– The runners are made of cast steel, cast iron or stainless steel

4. Draft tube – The pressure at the exit of the runner of a reaction turbine is generally less

than the atmospheric pressure

– The tail race is at atmospheric pressure. Thus water cannot be directly to tail race under normal conditions

– Thus a tube of increasing cross sectional area is used to make pressure difference by utilizing the KE of water. This tube is known as draft tube


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Degree of Reaction (R)

Degree of reaction (R) is defined as the ratio of pressure energy change inside a runner to the total energy change inside the runner.

Note : study the derivation for degree of reaction ( pg no 880. R.K.Bensal)


cot - cot 2

cot -1 R urbine,reaction t actualFor

0 R , binePelton tur For

runner theinsideenergy totalof Change

runner theinsideenergy pressure of Change R

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Inward radial flow turbines


• The water from the casing first enters the stationary guide vanes . From the guide vanes the water is directed to reach the runner which consist of moving vanes • The water flows over the moving vanes in the inward radial direction and is discharged at the inner diameter of the runner • The outer diameter of the runner is the inlet and the inner diameter of the runner is the outlet

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Velocity triangles, Work done and Efficiency of Inward radial flow reaction turbines


Hg

uV Therefore

Vf2V2 0,Vw1 and 90outlet at radial is discharge theif

Hg

uVuV

1000

HW1000

uVuV V a

or

VV a 2

1

uVuV V a

secper jet of KE

secper Workdone

secper jet of KE

secper Workdone , efficiency Hydraulic

diameterinner theis D ,60

NDu

diameterouter theis D ,60

NDu ,where

uVuVaV runner on secper Workdone

11wh

22w11w

22w11w1

h

2

11

22w11w1h

h

22

2

11

1

22w11w1

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Important formulas in radial flow reaction turbines


gH

uVuV

1000

HW1000

uVuVQ

efficiency Hydraulic.8

1000

uVuVQ (RP)runner on developedPower .7

uVuVg

1

2g

V-H

vaneshe through tflowsn water energy whe of loss no is thereIf 6.

VV 0,V and 0 meansoutlet at discharge Radial

0V and 0 meansinlet at discharge Radial .5

2g

V

g

PH Head .4

VB tn-DVB tn-DQ , turbineof Discharge

ion,considerat into taken are vanesof thicknessIf

flow of VelocityV

runner theofdiameter D

runner theof width B where

VBDVBDQ , turbineof Discharge .3

2gH

V ratio Flow .2

2gH

u ratio Speed .1

2w21w1

2w21w1

2w21w1

22w11w

2

2

2f2w2

w1

2

11

f222f111

f

f222f111

f1

1

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Outward radial flow turbines


• The water from the casing first enters the stationary guide vanes which is near to the shaft . From the guide vanes the water is directed to reach the runner which consist of moving vanes at the outer periphery of the shaft • The water flows over the moving vanes in the outward radial direction and is discharged at the outer diameter of the runner • The inner diameter of the runner is the inlet and the outer diameter of the runner is the outlet •For outward radial flow turbines, u1<u2 and D1<D2

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Numerical questions on outward radial flow turbines

• Refer prob 18.21 on pg no 893 (R.K.Bensal)

• Refer prob 18.22 on pg no 894 (R.K.Bensal)


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Francis turbine

• Francis turbine is an inward flow reaction turbine

• The discharge is radial at outlet. ie β=90° and Vw2=0 , Vf2=V2


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Francis Turbine

Francis turbine is a special type of inward flow reaction turbine

Refer page no 877 to 903 (RK.Bensal)


turbineflow radial inward of that as same as remains equations allRest

uVaV WD toreduces WorkdoneHence

VV and 0V ,90 Therefore

outlet.at radial is discharge the turbinefrancis In

1w1

2f2w2

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Velocity triangle of francis turbine


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Work done and Efficiency of Francis turbines


1

1

11w

11w1

h

2

11

11w1h

h

2121

22

2

11

1

11w1

D

B n , ratioBreadth

Hg

uV

1000

HW1000

uV V a

or

VV a 2

1

uV V a

secper jet of KE

secper Workdone

secper jet of KE

secper Workdone , efficiency Hydraulic

DD and uu here

diameterinner theis D ,60

NDu

diameterouter theis D ,60

NDu ,where

uVaV runner on secper Workdone

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fluid machinery module 1 - ktu notes

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