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HYDRAULIC TURBINES

1.Introduction :

2.Classification of Hydraulic Turbines :

Hydraulic (water) turbines are the machines which convert the water energy

(Hydro power) into Mechanical energy. The water energy may be either in the form of

potential energy as we find in dams, reservoirs, or in the form of kinetic energy in flowing

water. The shaft of the turbine directly coupled to the electric generator which converts

mechanical energy in to electrical energy. This is known as " Hydro-Electric power".

Water turbines are classified into various kinds according to i) the action of water

on blades, ii) based on the direction of fluid flow through the runner and iii) the specific

speed of the machine.

These may be classified into:1) Impulse type and 2) Reaction type

In impulse turbine, the pressure of the flowing fluid over the runner is constant and

generally equal to an atmospheric pressure. All the available potential energy at inlet will

be completely converted into which in turn utilized through a

purely impulse effect to produce work. Therefore, in impulse turbine, the available energy

at the inlet of a turbine is only the kinetic energy.In reaction turbine, the turbine casing is filled with water and the water pressure

changes during flow through the rotor in addition to kinetic energy from nozzle (fixedblades).As a whole, both the pressure and are available at the inlet ofreaction turbines for producing power.

Hydraulic machines are classified into :

a) Tangential or peripheral flow

b) Radial inward or outward flow

c) Mixed or diagonal flow

d) Axial flow types.

In tangential flow turbines, the water flows along the tangent to the path of rotationof the runner. Example:

In radial flow machine, the water flows along the radial direction and flow remainsnormal to the axis of rotation as it passes through the runner. It may be inward flow oroutward flow.

In Inward flow turbines, the water enters at the outer periphery and passes throughthe runner inwardly towards the axis of rotation and finally leaves at inner periphery.

(i) Based on the a

Francis turbine

ction of Water on Blades :

(ii) Based on the direction of Flow of Fluid through Runner :

a) Tangential Flow Machines :

Pelton wheel

b) Radial Flow Machines :

kinetic energy using nozzles,

energy kinetic energy

Example: . In outward flow machines the flow direction is opposite to theinward flow machines.

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

c) Mixed or Diagonal Flow :

Modern Francis turbine, Deriaz turbine

d)Axial Flow Devices :

(iii) Based on Specific Speed :

(a) Low Specific Speed:

(b) Medium Specific Speed :

(c) High Specific Speed :

In this type of turbine, the flow of fluid may enter at the outer periphery, passes overthe runner inwardly and leaves axially or parallel to the axis of rotation and vice-versa.

.

In this type of turbine, the water along the direction parallel to the axis ofrotation. Kaplan turbine, propeller turbine etc.,

Hydraulic turbines are classified into :

Which employs high head in the range of 200mup to 1700 m. These machines requires low discharge. Pelton wheel.N = 10 to 30 single jet and 30 to 50 for double jet Pelton wheel.

Which employs moderate heads in the range of 50m to200 m. Francis turbine, N = 60 to 400.

Which employs very low heads in the range of 2.5 m to 50m. These requires high discharge. Kaplan, Propeller etc., N = 300 to

1000.

This is a impulse type of tangential flow hydraulic turbine. It mainly posses:(i)Nozzle (ii) runner and buckets (iii) casing (iv) brake nozzle. Fig. 1 shows general layout ofhydro-electric power plant with pelton wheel.

The water from thedam is made to flowthrough the penstock. Atthe end of the penstock,nozzle is fitted whichconvert the potentialenergy into high kineticenergy. The speed of the jetissuing from the nozzle canbe regulated by operatingthe spear head by varyingthe flow area. The highvelocity of jet impingingover the buckets due towhich the runner startsrotating because of theimpulse effects and thereby hydraulic energy is converted into mechanical energy.After therunner, the water falls into tail race. Casing will provide the housing for runner and is opento atmosphere. Brake nozzles are used to bring the runner from high speed to rest conditionwhenever it is to be stopped. In order to achieve this water is made to flow in oppositedirection to that of runner .

Examples:

flowsExamples:

Examples:

Example:

Examples:

through brake nozzle

S

S

S

3 Pelton Wheel :

2

Head race

Tail race

Dam

Gross head(H )g

h

Penstock

SpearJet of water

Vanes

Casing

Nozzle

Net head(H)

Fig.1 Layout of Hydro-electric power plant

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

Turbine Generator

If = 1hhH

hhmechï ï

ïFig. 7.2 Various efficiencies of power plant

shaft output

Inlet

3

4.0 Heads and Efficiencies of Hydraulic Turbines :

4.1 Hydraulic Heads :

(a) Gross head:

(b) Effective head:

4.2

i) Hydraulic efficiency ( ):

___________________

ii) Volumetric efficiency ( ):

iii) Mechanical ( ):

iv) Overall efficiency ( ) :

It is the difference between the head race and tail race level whenthere is no flow.As such it is termed as static head and is denoted as H or H .

It is the head available at the inlet of the turbine. It is obtained byconsidering all head losses in penstock. If h is the total loss, then the effective

headabove the turbine is H = H - h

Various efficiencies of hydraulic turbines are:

i) Hydraulic ( ) ii) Volumetric ( )

iii) Mechanical ( ) iv) Overall ( )

It is the ratio of power developed by the runner to the water

power available at the inlet of the turbine.

= = =

It is the ratio of the quantity of water actually striking the

runner to the quantity of water supplied to the runner.

Q = amount of water that slips directly to

It is the ratio of shaft power output by the turbine to the

power developed by the runner.

=

It is the ratio of shaft output power by the turbine to the water

power available at inlet of the turbine.

s g

f

g f

vol

mech 0

vol

mech

mech

o

(1)

efficiency efficiency

efficiency efficiency

(2)

=.Where H is effective head at the inlet of turbine.

= = the tail race.

= loss. (3)

=

(4)

=

=

= ( )

= (5)

Efficiencies :

____________ ___

______

__ _

efficiency

___________________________

____________________

__________________

_______

h h

h h

h

rh

r

h

h

Dh D

h

h

r

h

h

hr

h h h h

h h h

H

H

H

H

vol

mech

o

o

o

o

h

(m /g ) (U V U V ) Q(U V U V ) H

(m gH ) / g Q g H H

H - h

H

Q Q - Q

Q Q

Shaft output power

Power developed by the runner

SP

Q (U V U V )/g

Shaft output power

Water power at inlet

SP

Q gH

.

.

c 1 u1 2 u2 1 u1 2 u2 e

c

f

1 u1 2 u2 c

vol mech

mech

± ±

±

a

th

H

H,act

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

5.0

W

V / 2g V / 2g

2

V

c

1 c 1 c

1

2 2

2

Work Done by the Pelton Wheel :

Where is the angle through which the jet is deflected by the bucket. is the

runner tip angle = 180 - .

Fig. 3 shows the inlet and outlet velocity . Since the angle of entrance of jet iszero, the inlet velocity triangle collapses to a straight line. The tangential component ofabsolute velocity at inlet V = V and the relative velocity at the inlet is V = V - U.

From the outlet velocity .

V = V cos – U

= V cos – U ( = V

V = (V – U) cos -U (

Work done / kg of water by the runner

W = U (V + V ) / g ( + ve sign for opposite direction of V and V )

= U [V + (V – U) c os – U] / g

= U [(V + (V – U) cos ] / g

W = U [(V – U) (1+C cos )] / g (6)

The energy supplied to the wheel is in the form of kinetic energy of the jet which is equal to

V /2g .

b

b

D

2

1 r1 1

u2 r2 2

b r1 2 r2 b r1

u2 b r1 1

u1 u2 c u1 u2

u1 b

1 1 b 2

1 c

c

1 b 2

1 b 2

triangles

C V C )

C V = V – U)

C

– U ) C

U [ (V – U) ( 1+C cos )] / gHydraulic efficiency, = =

U (V – U) ( 1+C cos )= (7)

u1

1 2

1 2 c

c

b 2

1

le

2

b

b

b

b

b

bh

bh

.

.

.

.

.

.

H

H

_______ _________________________

______________________

Fig 3 Shape of bucket & Velocity diagram

V1

Nozzle

=Vu1

qb2

b q2= 180 -0

U Vr1

U

Vr2

Vu2

V2Vf

ï ï ï

b2

4

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

V

(2U)

2

2

2

1

h

h

dFor maximum hydraulic efficiency, = 0

dU

2(1+ C cos )as , V – 2U = 0

V = 2U or = U/V = 0.5

(7.8)

2U (2U – U) ( 1+C cos )=

1+C cos= (9)

V = (10)

&Actual velocity of jet , V = C

C = coefficient of velocity for nozzle is in the range of 0.97 to 0.99

U = = U/( ), (11)

Where = Speed ratio and is in the range of 0.43 to 0.48

Total discharge, Q = n C (12)

Where n = number of jets (nozzles)

DN U 60Tangential velocity, U = or D = (13)

60 N

The ratio of mean diameter of buckets to the diameter of jet is known as"Jet Ratio". i.e., m = D/d.

m DZ = + 15 = + 15 where m ranges from 6 to 35. (14)

2 2d

h

b

\ f

b

b

b

Ö

Ö

f Ö Þ f Ö

f

Ö

pp

\

H

th

T

____

__________

________________________

___________

5.1 Working Proportions of Pelton Wheel :

(i) Ideal velocity of jet from the Nozzle,

( ii) Tangential velocity of buckets,

(iii) Least diameter of the jet, (d)

(iv) Mean Diameter or Pitch diameter of Buckets or Runner : (D)

_______ _______

(v) Number of Buckets required (Z) :

___ ____

b 21

1 1

1 1

b 2H, max

b 2H, max

2

1 v

v

v

U = V = 0.5 V

This shows that the tangential velocity of bucket should be half of the velocity of

jet for maximum efficiency.

Then,

If C = 1, then the above equation gives the maximum efficiency for = 0

½

b0

2gH

2gH

2gH 2gH

2gH

x

0

pd4

2pd4

2

V = n1

5

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

Example1 :

Solution :Given:

Total discharge Q :_________ _________

Diameter of wheel, D

Diameter of jet d,

Total discharge, Q

Number of Jets,

Example 2 :

Solution :Given:

Diameter of jet,

Pelton wheel has to be designed for the following data : power to be developed

= 5880 kW, Net head available = 300m, Speed = 550 RPM, ratio of jet diameter to wheel

diameter = 1/10 and overall efficiency = 85%. Find the number of jets, diameter of jet,

diameter of the wheel and the quantity of water required. Assume C = 0.98, = 0.46.

P= 5880 kW, H = 300m, N = 550 rpm, d/D= 1/10, = 0.85, C = 0.98,

= 0.46

Px 1000 Px 1000= or Q = = 2.35 m /s

Qg H gH

V =C = 0.98 =75.19 m/s

Tangential velocity of wheel, U= = 0.46 x = 35.29 m/s

U = DN/60

35.29 = x Dx 550/60

D = 1.225 m.

d/D = 1/10 d =D/10 = 1.225/10 = 0.1225 m

= n ( d /4)V

2.35 = n x0.1225 x75.19 /4

n = 2.65 3.

A single jet impulse turbine of 10 MW capacity is to work under ahead of 500m. If the specific speed = 10, over all efficiency = 0.8 and the coefficient

of velocity = 0.98, find the diameter of the jet and bucket wheel.Assume = 0.46.

P =10,000 kW, H =500m, N =10, = 0.8,C =0.98, d = ?, D = ?,= 0.46

0.8 = 10,000/(9.81 x 500 x Q)

Q = 2.548 m /s = d /4 V = d /4 x 97.06

d= 0.183 m or 18.3 cm

V

o V

1 V

1

1

OO

S O V

f

f

hh

Ö Ö

Ö Ö

»

h

r r

f

p

p

\

p

p

f

hf

p p

3

2gH 2 x 9.81 x 300

2gH 2 x 9.81 x 300

T

2

2

2 23

6

N = = = 10s_________ ____________

N = 236.4 rpm

Velocity of jet, V = C = 97.06 m/s1 V Ö2gH

N (P)

H

1/2

5/4

N

(500)

Ö10,000

5/4

Tangential velocity of bucket, U = = 0.46f Ö Ö2gH 2gH

U = 45.56 m/s

U = DN/60 = Dx236.4/60 = 45.56 m/sp pDiameter of a wheel , D = 3.68 m

= Px1000 / ( QgH)hO r

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

Example 3 :

Solution:

Given :

________ _____

Total flow,________ _____________

____ _______

Diameter of the jet,

Total tangential force,

A double jet pelton -wheel is required to generate 7500 kW when the

available head at the base of the nozzle is 400m. The jet is deflected through 165and the relative velocity of the jet is reduced by 15% in passing over thebuckets. Determine (i) the diameter of each jet (ii) Total flow (iii) Force exerted by

the jets in the tangential direction. Assume generator efficiency is 95%, = 80%, speed

ratio = 0.47

n =2, P =7500 kW, H =400, =15 = (180-165 ), V =0.85 V , =0.95

=0.8, =0.47=U/

Now, = (O/P)/(I/P) = P /P or P=P / = 7500/0.95 = 7894.74 kW

Out put by the turbine P = 7894.74 kW

Px1000 P= =

QgH QgH

P 7894.74Q = = = 2.515 m /s

gH 9.81x400x0.8

dV

2 x dAlso flow, Q = 2.515 = n = 88.6

4

d = 0.1344 m

F = Q (V - V ) /g

= 1000 x 2.515 (88.6 - 3.08)/1000

F = 215.08 kN

O

O O

h

b h

h f

h

hr

h

p p

4

r

O

O

1

g 2 r2 r1 g

o

g g g g

u1 u2 c

Ö

h

2gH

0

T

T

T

3

2 2

7

Velocity of jet from nozzle, V =

= = 88.6 m/s = V

1 Ö

Ö

2gH

2 x 9.81 x 400 u1

0.47=U/88.6 U= 41.64 m/sV = (V - U) = 88.6 - 41.64 = 46.96 m/s

V = 0.85 x 46.96 = 39.92 m/s

V =V sin 15 V =10.33 m/s

(U - V ) = V cos = 39.92 cos 15 = 38.56 m/s

V = 41.64 - 38.56

V = 3.08 m/s

Þ

Þ

r1

r2

f r2 f

u2 r2 2

u2

u2

1

0

0b

V = V1 u1

U Vr1

ï ï

ï ïïï

b2

b2

q=165o

Vr2

O/L

V2Vf

UVu2

ï ïï ï

TurbineP

G

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

6.0 Reaction Turbine :

In reaction turbines, only part of total head of water at inlet is converted intovelocity head before it enters the runner and the remaining part of total head is converted inthe runner as the water flows over it. In these machines, the water is completely filled in allthe passages of runner. Thus, the pressure of water gradually changes as it passes throughthe runner. Hence, for this kind of machines both pressure energy and kinetic energy areavailable at inlet. e.g., Francis turbine, Kaplan turbines, Deriaz turbine.

6.1 Francis turbine

8

Shaft

Scroll

casing

Scroll

casing

Link

Guide vane

Guide vane

Guide wheel

Runner

Fig. .4 Francis turbine

Regulating rod

Runner

Draft tube

Tail race

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

9

Francis turbine which is of mixed flow type is as shown in Fig. 4. It is of inwardflow type of turbine in which the water enters the runner radially at the outer periphery andleaves axially at its center.

This turbine consists of: (i)Scroll casing (ii)Stay ring (iii)Guide vanes(iv) Runner (v) Draft tube.

The water from penstock enters the scroll casing (called spiral casing)which completely surrounds the runner. The main function of spiral casing is toprovide an uniform distribution of water around the runner and hence to provideconstant velocity. In order to provide constant velocity, the cross sectional area of thecasing gradually decreses as the water reaching runner

The water from scroll casing enters the speed vane or stay ring. These arefixed blades and usually half in number of the guide vanes. Their function is to (a)direct the water over the guide vanes, (b) resist the load on turbine due to internalpressure of water and these load is transmitted to the foundation.

Water after the stay ring passes over to the series of guide vanes or fixedvanes. They surrounds completely around the turbine runner. Guide vanes functionsare to (a) regulate the quantity of water entering the runner and (b) direct the water onto the runner.

The main purpose of the other components is to lead the water to the runnerwith minimum loss of energy. The runner of turbine is consists of series of curvedblades (16 to 24) evenly arranged around the circumference in the space between thetwo plate. The vanes are so shaped that water enters the runner radially at outerperiphery and leaves it axially at its center. The change in direction of flow from radialto axial when passes over the runner causes the appreciable change in circumferentialforce which in turn responsible to develop power.

The water from the runner flows to the tail race through the draft tube.A draft is a pipe or passage of gradually increasing area which connect the exit of therunner to the tail race. It may be made of cast or plate steel or concrete. The exit end ofthe draft tube is always submerged below the level of water in the tail race and must beair-tight.

The draft tube has two purposes :(a) It permits a negative or suction head established at the runner exit, thus making it

possible to install the turbine above the tail race level without loss of head.

(b) It converts large proportion of velocity energy rejected from the runner into usefulpressure energy.

There are different types of draft tubes which are employed to serve the purpose inthe installation of turbine are as shown in Fig.5. It has been observed that for straight

divergent type draft tube, the central cone angle should not be more than 8 . This is

because, if this angle is more than 8 the water flowing through the draft tube withoutcontacting its inner surface which result in eddies and hence the efficiency of the drafttube is reduced.

(i) Scroll Casing :

(ii) Stay ring :

(iii) Guide Vanes :

(iv) Runner :

(v) Draft Tube :

7.0 Draft Tubes :

OO

mainly

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

(a) Straight divergent conical tube

(b) Moody spreading tube (or Hydracone)

(c) Simple elbow tube

(d) Elbow tube having circular cross section at inlet and rectangular cross section at outlet.

The absolute velocity at exit leaves the runner such that there is no whirl at exitie., V = 0. The Inlet velocity triangle are drawn for different conditions as shown in

Fig 6 (a) & (b).

8.0 Work Done and Efficiencies of Francis Turbine :

u2

10

Fig.5. Types of draft tubes

(d) Draft tube with circular inlet and rectangular outlet

(a) Conical drafttube

Tail race Tail race

Tail race

HsHs

Hs

(b) Simple elbowtube

(c) Moody Spreadingtube

U1U1

Vu 1 Vu 1

U1 D1aU2 D2a

V1

(a) When V > U

or < 90O

u 1 1

1b

(b) If When U

or > 90O

1 u 1

1

> V

b

Radial discharge

ie., V = V

& = 90

or V

O

2 f2

2

u2 = 0

a

V1

Vr1

Vf1Vr1

Vf1

ï ï ï

ï

U2 ïï

ïïï

ï

Outlet

Outlet velocity triangle

a2 b2

Vr2V2 f2=V

D2

Runnerblade

D1

a1 b1 a1b1

Runner U2 ïï

b2

Vr2

Vu 2=0

V2=Vf2

w

Vu 1

V1

Vr1Vf1

ïï

InletInletInlet

b1a1

Fig.6 Velocity triangles for different conditions

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

(i) Work done / kg of water ,

(ii) Hydraulic efficiency :

________________________

______

_____________ =______

___________ __________

(iii) Volumetric efficiency ( ):

__________

(iv) Mechanical efficiency ( ):

____________________________

(v) Overall efficiency ( ):

_________________________ ________________

8.1 Working Proportions of Francis Turbine :

flow ratio

area of flow

W = (U V - U V )/g

As V = 0 in francis turbine (axial outlet velocity)

W = U V /g (15)

Work done by the runner=

Available energy at inlet

U VQ g H

=QgH Hg

Also gH - (V / 2g ) H - (V /2g)= = (16)

gH H

Where H is the head at inlet, V = velocity of water at exit of the draft tube

It is the ratio of quantity of water through the runner

(Q ) to the quantity of water suppled (Q )

(Q - Q)= Q / Q =

Q

Q = loss = Q - Q

Power output at the shaft end=

Power developed by the runner

Power output at the shaft Shaft power= =

Power available at inlet of the turbine Water power at inlet

= ( x ) = x (17)

(1) Tangential velocity of runner (U ):

Speed ratio, = U / , where ranges from 0.6 to 0.9 (18)

(2) The ratio of flow velocity V at inlet tip of the vane to the spouting velocity ( )is known as the

= V / , where ranges from 0.15 to 0.30 (19)

(3) If n is the number of vanes in the runner, 't' is the thickness the vane at inlet

and B is the width of the wheel at inlet then, the at inlet,

A = ( D - n t)B = C D B (20)

1 2

1

H

1

H

H

4

2 1

1vol 2 1

1

1 2

mech

o

o H vol mech H,act mech

1

1

f1

f1 1 1

u1 u2 c

u2

u1 c

u1

c

c

c 4

f1

h

h

rh

r

h

h

Dh

D

h

h

h

h

h h h h h h

f Ö f

Ö

\ y Ö y

p p

H

vol

mech

o

e

2 2

4

2gH

2gH

2gH

1

1

11

[ [

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

where C = Contraction factor which represents the area occupied by the vane thickness

in the runner and it is a few percentage of D .

Discharge, Q =A V = C D B V = C D B V

Normally it is assumed that D = 2D & V = V , B = 2B

Hydraulic efficiency is given by = if the velocity V the exit of

draft tube is given since V /2g represents the head loss at the exit of draft tube.

Actual regain of pressure head=

Velocity head at entrance to draft tube.

(V - V )/2g -h (V - V )/2g -h= = (21)

V /2g V /2g

Where V = Velocity of water at exit of runner = V

V = Velocity of water at exit of draft tube.

h = head lose in draft tube.

. Show that the utilization factor for an inward flow reaction turbine withrelative velocity component at inlet perpendicular to the tangent of the wheel and the

absolute velocity at the exit is radial is given by = 2 cos / (1+ cos )

Where is the angle made by the entering fluid with tangent of the whee.

Utilization,

= = =

= = ( V = V - U )

= =

p

p p

h

h

[ ] [ ]h

Î a a

a

Î

Î

1

f 1 1 1 2 2 2

1 2 1 2 2 1

4

4

d

2 4 f 3 4 f

d

2 3

3

4

f

1 1

1

2 1 1

f f f

f f

2

Note :__________

Efficiency of the draft tube :

_______________________________

________________ ________________

Example 4 :

Solution :

____________ ______________ _______________

____________ _________

____________ ____________

H2

2 2 2 2

2 2

2 2

2 2 2.

.

.

12

H- (V /2g)

H

24

V = V2 3

V1

V4

3

2

4

Combined Inlet & outlet velocity tri nglea

V = 0

VV

V

V

U = V Uï ï ï

a b

b a=

ba

E

E + V /2g2

2 c

U V

U V + V /2g

1

1 2

u1

u1 c2

U

U + V /2

2

2 2

1

1 2

2U

2U + V -U

2

2 2 2

1

1 1 1

2U

U + V

2

2 2

1

1 1

2V cos

V cos + V

2 2

2 2

1 1

1 1 1

a

a2

2 cos

1+ cos

2

2

a

a1

1

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

13

Example 5 :

Solution : Given :

(i) Euler's head :

(ii) Power developed (P) :

(iii) Inlet blade angle ( ):

(iv) Degree of reaction (R)

_________________ ___________________________

In a Francis turbine, the discharge is radial. The blade speed at inlet = 25 m/s.At the inlet tangential component of velocity = 18 m/s. The radial velocity of flow is

constant and equal to 2.5 m/s. Water flows at the rate of 0.8 m /s. The utilization factor is0.82. Find i) Euler's head ii) Power developed iii) Inlet blade angle iv) Degree ofreaction (R). Draw the velocity triangles.

U = 25 m/s, V =18 m/s, V = 2.5 m/s, Q = 0.8m /s,

gH = E = (U V ) /g as V = 0,

E = 25 x 1 8/1 = 450 J/kg = gH

H = 450/9.81 = 45.87 m.

P= QE = Q (gH )= (1000 x 0.8 x 450) / 1000

P= 360 kW

V = (V +V ) = 18.17 m/s V =V = V = V = 2.5 m/s

tan = V /(U - V ) = 2.5 / (25 - 18)

= 19.65

H - [ (V - V )]/2g 45.87 - [(18.17 - 2.5 )/2 x 9.81]R = =

H 45.87

R = 0.64

3

:

1 f

e 1

e

e

e

1 f u 1 f1 f 2 f 2

1 f 1 u1

1

1 2

u1

u1 c u2

3

\

r r

b

b

b

1

2 2 1/2

2 2

0

2 2

e

e

OutletInlet

V

V = 0

ïï

ï

Vr1Vr2V1

V2Vf

=Vf2

U

U

ï ïï

a1 a2b1 b2

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

9.0 Propeller & Kaplan Turbines

The propeller turbine consists of an axial flow runner with 4 to 6 blades of aerofoilshape. The spiral casing and guide blades are similar to that of the Francis turbines. In thepropeller turbines, the blades mounted on the runner are fixed and non-adjustable. But inKaplan turbine the blades can be adjusted and can rotate about the pivots fixed to theboss of the runner. This is only the modification in propeller turbine. The blades areadjusted automatically by a servomechanism so that at all loads the flow enters withoutshock.

The Kaplan turbine is an axial flow reaction turbine in which the flow is parallel tothe axis of the shaft as shown in the Fig.7.7. This is mainly used for large quantity of waterand for very low heads (4-70 m) for which the specific speed is high. The runner of theKaplan turbine looks like a propeller of a ship. Therefore sometimes it is also called aspropeller turbine. At the exit of the Kaplan turbine the draft tube is connected to dischargewater to the tail race.

1) At inlet, the velocity is asshown in Fig.8.

2) At the outlet, the discharge isalways axial with no whirlvelocity component i.e., outletvelocity triangle just a right angletriangle as shown Fig.8.

Dle

14

Fig.7.7 Main components of Kaplan turbine.

Draft tube

Tail raceBoss

Guide vane Guide vaneScroll casing

Inlet of runner vanesRunner vanes

dD

Shaft

Outlet of runner vanes

Fig.8 Velocity triangles for Kaplan Turbine.

ïï

Vf =V2

Vu 2=0

Vr2

U

b

Inlet Outlet

V1

Vf

Vr1

U

Vu 1

ïï

ïï

a b1

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

9.1 Working Proportions of Kaplan Turbine

i) Tangential velocity of blades based on tip diameter (U )

ii) Flow velocity (V )

iii) The discharge (Q) flowing through the runner is given by

____ ____

Example 6 :

Solution :

___ ____

____

D = 5.91 m

N = 114.5 rpm

Example 7:

D

U =

where, D = Tip diameter or Outer diameter of the runner

d = Hub diameter or Boss diameter of the runner

d/D = 0.35 to 0.60

= V / Usually, 0.35 < < 0.75

Q = (D - d )V = (D - d )4 4

:

Akaplan turbine produces 80,000 HP(58,800 kW) under a head of 25m which

has an overall efficiency of 90%. Taking the value of speed ratio = 1.6, flow ratio = 0.5and the hub diameter = 0.35 times the outer diameter. Find the diameter and the speed of theturbine.

P = 80,000 HP = 58,800 kW, H = 25m, = 0.9, = 1.6, = 0.5,

d/D =0.35, :D, N.

= P/( QgH)

Q = P/( gH) = 58,800 / (0.9 x 9.81 x 1 x 25)

Q = 266.4 m /s

Again, Q = (D - d ) = D 1- (d/D) x4 4

266.4 = D (1 - 0.35 )x (0.5)4

Also, = U = U / = 1.6

U = 35.44 m/s = DN/60 = x 5.91 x N / 60

.

Determine the efficiency of a Kaplan turbine developing 2940 kW under ahead of 5m. It is provided with a draft tube with its inlet diameter 3m set at 1.6m above thetail race level.Avacuum pressure gauge connected to the draft tube inlet indicates a readingof 5m of water.Assume that draft tube efficiency is 78%.

D f Ö

y Ö y

y Ö

f y

h f y

h r

h r

y Ö y Ö

Ö

f /Ö Öp p

\

2gH

2gH

2gH

2gH 2gH

2gH

(22)

(23)

(24)

\

p p

p p

p

f

f

f

2 2 2 2

3

2 2 2 2

2 2

Note Unless otherwise mentioned, the velocity triangles will always be drawn w.r.t tipdiameter.

OTo find

0

0

2 x 9.81 x 25

(2 x 9.81 x 25)D D

D

15

[ ]

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri

Solution :Given :

___________________________

____________ __________

____

__________________ ________

_____________________

P = 2940 kW, H = 5 m, H =1.6 m, p / g = - 5 m, = 0.78, = ?

Apply the Bernoulli's equation between (2) & (3) we get

p / g = - H + (V - V )/2g

- 5 = - 1.6 - (V - V )/2g

(V - V )/2g = + 3.4 m

(V - V ) = + 66.708 (1)

Draft tube efficiency =

(V -V )/2g (V - V )= = = 0.78

V /2g V

1- (V /V ) = 0.78

V = 0.22 V (2)

n equation (1), V (1 - 0.22) = 66.708

V = 9.25 m/s = V ( Axial discharge in Kaplan turbine)

Now discharge, Q =A V = x 3 x 9.25 = 65.38 m /s4

Overall efficiency of a Kaplan turbine,

= =

= = 91.68 %

s 2 draft o

2 s

d

f

f f

0

0

r h h

r

h

\

\ I

hr

2 3

2 3

2 3

2 3

2 3 2 3

2 2

3 2

3 2

2

2

2 2

2 2

2 2

2 2

2 2 2 2

2 2

2 2

2

Actual pressure gain in draft tube

Pressure at inlet to the tube.

Power output P

Hydro power at inlet QgH

2940 x 1000

1000 x 9.81 x 65.38 x 5

2

2 3

.

.

.

p

h

16

2

3

p / g =52 r

Datum line

d = 38 m2

2

Z1

Z2

Hs

Hs =1.6 m

[ ]

[ ]

Dr.M.S.Govinde Gowda, Principal, Alva’s Inst. Of Engg & Tech, Moodbidri