FLUID MACHINES 1 Introduction

The device which converts hydraulic energy into mechanical energy or vice versa is known as

Fluid Machines.

Since a rotating element called ROTOR is an essential part of these machines, and since mutual

dynamic action between the rotor and the working fluid forms their basic principles of operation,

these are also called ROTODYANAMICS or TURBOMACHINES.

The hydraulic machines which convert hydraulic energy into mechanical energy are known as

Turbines and that convert mechanical energy into hydraulic energy are known as Pumps.

Fig.1 shows a general layout of a hydroelectric plant.

Fluid Machines/ Rotodynamic machines/Turbo Machines

Incompressible Fluid (Water) Compressible Fluid (Steam, gas, air)

Hydrodynamic Machines

Fluid to rotor Rotor to Fluid

Hydro-Turbines

Hydrodynamic pumps

Thermal Turbo-machines

Fluid to rotor Rotor to fluid

Steam and gas turbines

Compressor or blower

Fig. 1 General layout of hydropower plant

It consists of the following:

A Dam constructed across a river or a channel to store water. The reservoir is also known as

Headrace.

Pipes of large diameter called Penstocks which carry water under pressure from storage reservoir

to the turbines. These pipes are usually made of steel or reinforced concrete.

Turbines having different types of vanes or buckets or blades mounted on a wheel called runner.

Tailrace which is channel carrying water away from the turbine after the water has worked on

the turbines. The water surface in the tailrace is also referred to as tailrace.

2. Important Terms

(i) Gross Head (Ha): It is the vertical difference between headrace and tailrace.

It is the total specific energy of liquid on pressure side of a machine with respect to specific

energy in tail race or sump as datum. Thus with reference to Fig. 2,

s

fooa

hg

VVpH

2

22

(1)

where po is intensity of pressure on pressure side of machine, Vo is velocity of liquid on pressure

side of machine, Vf is the velocity at tail race, hs is the height of machine from sump or tail race

level generally measured up to centre line of machine.

Fig. 2 Head and Discharge

(ii) Net Head (H): Net head or effective head is the actual head available at the inlet of the

turbine.

LahHH (2)

Where hL is the total head loss during the transit of water from the headrace to tailrace. This

mainly consists of head loss due to friction given by:

gd

flVh

f 2

42

(3)

Here, f is the coefficient of friction of penstock depending on the type of material of penstock, L

is the total length of penstock, V is the mean flow velocity of water through the penstock, D is

the diameter of penstock and g is the acceleration due to gravity.

(iii) Discharge

The discharge (Q) through the machine is expressed in cubic meter per second (m3/s) or

litres/second (l/s).

In machines there has to be a gap between stationary and rotating parts. A part of discharge does

not enter the runner at all and instead passes through this gap to the tail race without doing any

work.

Thus if Q1 is the quantity of liquid available on pressure side of machine, the net quantity of

water, Q, doing work in turbine will be (Q1 - Q) where Q is the amount of discharge which

leaks away without doing any work.

In pumps although the actual quantity of liquid lifted is Q+Q, only Q is supplied at outlet and

Q leaks back to sump.

The amount of energy, E, released when an object of mass m drops a height H in a gravitational

field of strength, g, is given by

mgHE

The energy available to hydroelectric dams is the energy that can be liberated by lowering water

in a controlled way. In these situations, the power is related to the mass flow rate, Q.

QHQgHgHt

mgH

t

m

t

E

N-m/s

is specific weight = 81.91000 N/m3 in SI units.

So that power generated, QHP N-m/s QH Watt (4)

And water horse power to the turbine, 75

...QH

PHW

metric H.P= 13.33 QH mHP

= 736.033.13 QH = 9.8QH [kW]

Electric energy = [kW] = QH8.9 = QH8 kW (5)

Taking efficiency equal to 80%.

3. Types of efficiencies

Depending on the considerations of input and output, the efficiencies can be classified [Fig.3] as:

1. Leakage or Volumetric Efficiency

2. Hydraulic Efficiency

3. Mechanical Efficiency

4. Overall efficiency

Fig. 3 Elements of a simple turbine

Considering turbines and pump separately, we may examine these efficiencies for a turbine as

follows:

3.1 Volumetric Efficiency:

Let a hydraulic turbine be supplied discharge (Q) at the head (H).

Then water horse power (WHP) supplied to the machine is given by

75...

QHPHW

(5)

However, due to leakage of Q amount of water to tail race without doing useful work, power

wasted in leakage is given by

Working horse power wastage = 75

QH (6)

Hence volumetric efficiency is given by:

QQQ

Eq

EqEqv

)3.(

)4.()3.( (7)

3.2 Hydraulic Efficiency

It is the ratio of the power developed by the runner of a turbine to the power supplied at the inlet

of a turbine. Since the power supplied is hydraulic, and the probable loss is between the striking

jet and vane it is rightly called hydraulic efficiency.

If R.P. is the Runner Power and W.P. is the Water Power, then

PW

PRh .

. (8)

Net power supplied to turbine = 75

)( HQQ (9)

In the runner specific energy conversion is Hr and hence hydraulic power generated by the

runner is given by

75

)(r

H

HQQP

(10)

So that hydraulic efficiency is given by:

H

H

HQQ

HQQ

r

r

h

75

)(75

)(

(11)

3.3 Mechanical Efficiency

It is the ratio of the power available at the shaft to the power developed by the runner of a

turbine. It depends on the slips and other mechanical problems that will create a loss of energy

between the runner and shaft which is purely mechanical and hence mechanical efficiency.

If S.P. is the Shaft Power

PR

PSm .

. (12)

If Pm represents loss of power in mechanical friction at bearings etc., the power available at the

shaft (Ps) will be obtained by subtracting Pm from hydraulic power (PH) produced. Thus

Ps = PH - Pm (13)

So that the mechanical efficiency is given by

sm

s

H

s

mPP

P

P

P

(14)

The effect of disc friction losses is similar to that of mechanical friction and hence is included

into it.

3.4 Overall Efficiency

It is the ratio of the power available at the shaft to the power supplied at the inlet of a turbine. As

this covers overall problems of losses in energy, it is known as overall efficiency.

It depends on both the hydraulic losses and the slips and other mechanical problems that will

create a loss of energy between the jet power supplied and the power generated at the shaft

available for coupling of the generator.

PW

PSo .

. (15)

vhm

r

mH

mmsms

o QH

HQQ

HQQ

HQQ

PHW

P

PHW

PP

PHW

P

75

75

)(

75

)(75

)(

......

)(

..

(16)

Since the mechanical losses are considered as outside effects on turbine performance, hence term

hvi indicating internal efficiency is also sometimes used.

Preceding in the same manner the above efficiencies can be derived for pumps also.

4 Classifications of Turbines

4.1 Based on the action of water on the blades

(a) Impulse Turbines:

In this, water is brought to the turbine through a pipe called penstock at the end of which a

nozzle is provided to convert pressure energy into kinetic energy before it strikes the blades in

the atmosphere, hence also sometimes called free jet turbines. The turbine is enclosed in a casing

to prevent splashing of water. Otherwise, it has no other hydraulic function. Example: Pelton

Turbines [Fig. 4].

Fig. 4 Pelton Turbine

(b) Reaction Turbines

Here water enters the turbine under pressure and hence has both pressure energy and kinetic

energy. This turbine always runs full. Examples: Francis turbine [Fig. 5], Kaplan turbine [Fig. 6],

Propeller turbine.

Fig. 5 Francis turbine

Fig. 6 Kaplan Turbine

Impulse turbines are used for high heads and low discharges while reaction turbines for low head

and high discharge.

4.2 Based on the direction of flow

(a) Radial flow turbine

Here path of water through runner lies in a plane perpendicular to the axis of rotation. It can be:

(i) Outward radial flow turbine: Water enters at the inner circumference and discharges at the

outer circumference of the runner.

(ii) Inner Radial flow turbine: Here water enters at outer and leaves at inner circumference.

(b) Axial Flow Turbine

Water enters parallel to the shaft e.g. Kaplan turbine

(c) Tangential flow turbine

Water enters tangentially. Example: Pelton Turbine

(d) Mixed flow turbine

Water enters radially at the outlet and leaves in the axial direction at the inlet of the runner e.g,

Francis turbine

4.3 Based on the relative position of shaft

(a) Horizontal Shaft turbine

Here shaft of the turbines lies in horizontal plane. Example: Pelton turbine

(b) Vertical Shaft turbine

Turbines producing greater power normally have their shafts vertical e.g. Francis, Kaplan.

They can be arranged as per the following table:

As can be seen from the above table, any specific type can be explained by suitable construction

of sentences by selecting the other items in the table along the row.

5. Pelton wheel or turbine

Pelton wheel [Fig. 7], named after an eminent engineer, is an impulse turbine wherein the flow is

tangential to the runner and the available energy at the entrance is completely kinetic energy.

Further, it is preferred at a very high head and low discharges with low specific speeds. The

pressure available at the inlet and the outlet is atmospheric.

Turbine Type of

energy Head Discharge

Direction

of flow

Specific

Speed Name Type

Pelton Wheel Impulse Kinetic

High

Head > 250m

to 1000m

Low Tangential

to runner

Low

Fig. 7 Pelton Wheel

The main components of a Pelton turbine are:

5.1 Nozzle and flow regulating arrangement

Water is brought to the hydroelectric plant site through large penstocks at the end of which there

will be a nozzle, which converts the pressure energy completely into kinetic energy. This will

convert the liquid flow into a high-speed jet, which strikes the buckets or vanes mounted on the

runner, which in-turn rotates the runner of the turbine. The amount of water striking the vanes is

controlled by the forward and backward motion of the spear [Fig. 8]. As the water is flowing in

the annular area between the nozzle opening and the spear, the flow gets reduced as the spear

moves forward and vice-versa.

Fig. 8 Flow control for Pelton turbine

5.2 Runner with buckets

Runner is a circular disk mounted on a shaft. On its periphery, a number of buckets are fixed

equally spaced as shown in Fig 9.

Fig. 9 Runner of a Pelton turbine

The buckets which may be hemispherical or double ellipsoidal [Fig. 10], are made of cast-iron

cast-steel, bronze or stainless steel depending upon the head at the inlet of the turbine. The water

jet strikes the bucket on the splitter of the bucket and gets deflected through 160 - 1700. The

shape is so designed to get maximum possible energy from the jet.

Fig. 10 Detail of bucket

5.3 Casing

It is made of cast-iron or fabricated steel plates [Fig. 7]. The main function of the casing is to

prevent splashing of water and to discharge the water into tailrace.

5.4 Breaking jet

Even after the amount of water striking the buckets is completely stopped, the runner goes on

rotating for a very long time due to inertia [Fig. 7]. To stop the runner in a short time, a small

nozzle is provided which directs the jet of water on the back of bucket with which the rotation of

the runner is reversed. This jet is called as breaking jet.

5.5 Deflector

In the path of the jet near the nozzle a deflector is provided to control the quantity of the jet

striking the bucket.

5.6 Tail Race

It is kept at such a level that discharge comes out of turbine under gravity and at atmospheric

conditions.

Small turbines are cast as a single unit while in large turbines buckets are cast separately and

bolted to the runner. This facilitates repairs or parts displacement.

6. Working properties of Pelton Wheel (Turbines)

1. Ideal velocity or theoretical velocity of jet is called sprouting velocity = gH2 (17)

Due to friction of nozzle, actual velocity of jet, gHCVv

2 . Here Cv varies from 0.97 to 0.99.

2. The maximum velocity of wheel u = 0.5V to get maximum efficiency. In actual practice:

gHkgHgHVuu

2245.0298.046.046.0 (18)

Here ku varies from 0.43 to 0.47

3. Least diameter of jet, 2

121

542.02

4

H

Q

gHC

Qd

v

(19)

d is in m when Q is in m3/s.

4. Mean diameter of Pelton wheel is called Pitch dia.

60

DNu

N

gHk

N

uD u

)2(6060 (20)

5. The ratio of mean or Pitch diameter of wheel to the diameter of jet is called jet ratio and is

denoted by m = (D/d). Here, m varies from 11 to 14 and an average value of 12 is generally

opted.

Smaller values of m result in too close spacing or too few buckets for whole jet. Large value

results in too bulky installation.

6. Dimensions of bucket and jet

(a) Axial width, B = (3 to 4) d

(b) Radial length, L = (2.5 to 3.0)d

(c) Depth of cup, y = (0.8 to 1.2)d

(d) Width of cut, w= (1.5 to 2.0)d

(e) Length of cut, L1 = (0.3 to 0.5)d

(f) No. of buckets = 152

1

d

D or

21

4.5

d

D (21)

(g) Area of jet < 0.1 (Projected area of bucket, =L*B)

(h) Angle of deflected jet, = 160o to 170o

Cut in the bucket is provided so that the bucket exactly opposite the jet gets full jet striking on it

and is not intercepted by the lower portion of the bucket that follows.

Number of buckets should be as few in number as possible so that friction loss is minimum and

as maximum as possible so that water is fully utilized and no water goes waste.

7. Theory of Pelton wheel

Figure 11 shows 3 D picture of a jet striking the splitter and getting split in to two parts and

deviating.

Fig. 11 Jet striking pelton wheel blade

Figure 12 shows velocity triangle for the jet striking pelton wheel bucket.

Fig. 12 Velocity triangles

u Peripheral velocity at the center of the bucket

V1 inlet absolute velocity

Vr1 relative velocity at the inlet = (V1 u)

Vr2 relative velocity at the outlet = Vr1 = (V1 u)

V2 absolute velocity at the outlet

Vw2 velocity of whirl at the outlet

Vf2 velocity of flow at the outlet

and angles with which jet enters and leaves the turbine

vane angle at the inlet

vane angle at the outlet also called side clearance angle

The ratio of peripheral velocity, u, to the inlet absolute velocity, ( )21

gHKVv

is called

VELOCITY RATIO (KV).

1V

uK

v

)cos(21

VVQFx

(22)

Now

])cos1(cos[}cos){()cos(coscos111222

uVuuVuVuVVVrrw

so that )cos1)((])cos1(cos[111

uVQuVVQFx

Work done per second i.e. power = )cos1)((1

uVQuuFx

(23)

0)cos1)(2(1

uVQdu

dP

11 5.0

2V

Vu (24)

Hence 4

)cos1(

2

1max

VQP (25)

Hydraulic Efficiency,

cos112

2

)cos1)((

112

1

1

V

u

V

u

VQ

uuVQ

h (26)

Mechanical Efficiency, uuVQ

P

jet to due shaft on acting Power

shaft at Available Power sm )cos1)((

1

(27)

Overall efficiency, mh

For Pelton wheel, %9995m and %9085 h

7. Governors for Turbines

Generators driven by turbine are to run at a constant speed to produce power at a given

frequency irrespective of power requirement or load. The frequency is given by

60

pNf (28)

Here N is the r.p.m. of generator; p is the number of pairs of poles for generator

Usually, f = 50, so that pp

N30006050

(29)

Hence, it is necessary that turbine should also run at constant speed N at all loading conditions.

This speed is known as SYNCHRONOUS SPEED and the turbines are designed for this speed.

The Pelton turbine speed can be maintained constant:

(i) by controlling discharge by forward and backward motion of spear inside the nozzle

(ii) by deflecting the discharge using deflector, or

(iii)by combination of these two

This regulation of discharge flowing through the runner in accordance with the variation in load

is called governing of the turbine and usually this is done automatically by means of a

governor.

A Governor is a mechanism to regulate the speed of the shaft of a turbine.

The most commonly used governor used is oil pressure governor as shown in Fig. 13.

Fig. 13 Oil pressure governor

The main component parts of the governor are:

1. Servo-meter or relay cylinder

2. Relay valve or control valve or distribution valve

3. Actuator or pendulum driven by belt or gear connected to the main shaft of turbine

4. Oil Sump

5. Oil pump driven by belt connected to the turbine main shaft.

6. System of oil supply pipes connecting oil sump with relay valve and relay valve with

servo-meter.

Working

Fig. 14 Working of a Governor

With respect to Fig. 14

When load on generator drops, speed of generator and hence turbine increases.

This result in outward movement of fly balls, hence the upward movement of sleeve.

This result in downward movement of lever on R.H.S of fulcrum and hence downward

movement of Bell crank lever.

This downward movement will raise the deflector up, coming in the way of jet, deflecting it

away from the vane.

The downward motion of main lever, also move the piston in the control or relay valve

downward which will result in opening servo-meter on the left hand side of its position which

then will move to the right side.

This causes the spear to move forward. This will reduce the area of the nozzle outlet and thus

decreases the rate of flow and normal speed is restored.

When normal speed is restored, main lever will attain its normal position and the roller on the

cam will move up forcing the belt crank to move up and hence the deflector will come to its

original position.

8 Some other Types of Impulse Turbines

(i) Double overhung Pelton Wheel Turbine

When high speed or greater power is required, the two pelton wheels are provided on a single

shaft. However, analysis will remain same as for single pelton turbine.

(ii) Multiple Jet Pelton Turbine:

The power developed by a single wheel may be increased by providing more than one jet [Fig.

15] spaced evenly around the runner. The jet should be so spaced that the water issued from a jet

after striking the runner should not interfere with other jets. The maximum number of jets so far

used in some larger units is six. The power of such wheel will be (n X P) where n is the number

of jets and P is the power due to single jet.

Fig. 15 Muti-jet Pelton Turbines

9. Reaction Turbines

Reaction turbines are those turbines which operate under hydraulic pressure energy and part of

kinetic energy. In this case, the water reacts with the vanes as it moves through the vanes and

transfers its pressure energy to the vanes so that the vanes move and in turn rotate the runner on

which they are mounted.

9.1 Types of reaction turbine

The main types of reaction turbines are:

(i) Radially outward flow reaction turbine: This reaction turbine consists of a cylindrical disc

mounted on a shaft and provided with vanes around the perimeter. At inlet the water flows into

the wheel at the centre and then glides through radially on fixed guide vanes and then flows over

the moving vanes. The function of the guide vanes is to direct or guide the water into the moving

vanes in the correct direction and also regulate the amount of water striking the vanes. The water

as it flows along the moving vanes will exert a thrust and hence a torque on the wheel thereby

rotating the wheel. The water leaves the moving vanes at the outer edge. The wheel is enclosed

by a water-tight casing. The water is then taken to draft tube.

(ii) Radially inward flow reaction turbine: The constitutional details of this turbine are similar

to the outward flow turbine but for the fact that here the guide vanes surround the moving vanes.

This is preferred to the outward flow turbine as this turbine does not develop racing. The

centrifugal force on the inward moving body of water decreases the relative velocity and thus the

speed of the turbine can be controlled easily.

(iii) Mixed flow reaction turbine: This is a turbine wherein it is similar to inward flow reaction

turbine except that when it leaves the moving vane. The direction of water is turned from radial

at entry to axial at outlet. The rest of the parts and functioning is same as that of the inward flow

reaction turbines.

(iv) Axial flow reaction turbine: This is a reaction turbine in which the water flows parallel to

the axis of rotation. The shaft of the turbine may be either vertical or horizontal. The lower end

of the shaft is made larger to form the boss or the hub. A number of vanes are fixed to the boss.

When the vanes are composite with the boss the turbine is called propeller turbine. When the

vanes are adjustable the turbine is called a Kaplan turbine.

Fig. 16 Inward radial flow reaction turbine

9.2 Derivation of the efficiency of a reaction turbine

Consider Fig. 17.

Fig. 17 Reaction turbine velocity triangles

Let

R1 = Radius of wheel at inlet of the vane

R2 = Radius of wheel at outlet of the vane

= Angular speed of the wheel

Tangential speed of the vane at inlet = u1 = R1

Tangential speed of the vane at outlet = u2 = R2

The velocity triangles at inlet and outlet are drawn as shown in Fig. 17.

and are the angles between the absolute velocities of jet and vane at inlet and outlet

respectively

and are vane angles at inlet and outlet respectively

The mass of water striking a series of vanes per second = a V1

Here a is the area of jet or flow and V1 is the velocity of flow at inlet.

The momentum of water striking a series of vanes per second at inlet is given by the product of

mass of water striking per second and the component of velocity of flow at inlet

= a V1 x Vw1 (Vw1 is the velocity component of flow at inlet along tangential direction)

Similarly momentum of water striking a series of vanes per second at outlet is given by

= a V1 x (Vw2) (Vw2 is the velocity component of flow at outlet along tangential direction and

with sign because the velocity component is acting in the opposite direction)

Now angular momentum per second at inlet is given by the product of momentum of water at

inlet and its radial distance = a V1 x Vw1 x R1 (30)

And angular momentum per second at inlet is given by = a V1 x Vw2 x R2 (31)

Torque exerted by water on the wheel is given by impulse momentum theorem as the rate of

change of angular momentum:

)22111221111

R x Vw R x Vw ( V a =)]R x Vw x V a (- - R x Vw x V a [ = T (32)

Work done per second on the wheel = Torque x Angular velocity = T x

WD/s = a V1 (Vw1 R1+ Vw2 R2) x = a V1 (Vw1 R1 x + Vw2 R2 x ) (33)

As u1 = R1 and u2 = R2, we can simplify the above equation as

WD/s = a V1 (Vw1 u1+ Vw2 u2) (34)

In the above case, always the velocity of whirl at outlet is given by both magnitude and direction

as: Vw2 = (Vr2 Cos u2) (35)

If the discharge is radial at outlet, then Vw2 = 0 and hence the equation reduces to

WD/s = a u1V1 Vw1 (36)

KE/s = a V13 (37)

Efficiency of the reaction turbine is given by

3

1

22111

21 Va

uVuVVa

second / EnergyKinetic

second / done Work ww

2

1

22112

V

uVuV ww (38)

Note: The value of the velocity of whirl at outlet is to be substituted as Vw2 = (Vr2 Cos u2)

along with its sign.

Work done per unit weight = second per striking water of Weight

second per done Work

=

2211

2211 1 uVuVggQ

uVuVQ

ww

ww

(39)

If the discharge at the exit is radial, then Vw2 = 0

Work done per unit weight = 11

1uV

g w (40)

Hydraulic efficiency =

2211

2211 1

..

..uVuV

HgHQg

uVuVQ

PW

PRww

ww

(41)

If the discharge at the exit is radial, then Vw2 = 0 and hence

Hydraulic efficiency, = 11

1uV

Hg w (42)

Fig. 18 Velocity Triangles for different reaction turbines

9.3 Francis Turbine

The inward flow reaction turbine having radial discharge (i.e. the angle made by the absolute

velocity with the tangent on the wheel is 90o) at the outlet is known as Francis turbine. The main

component parts of a Francis turbine [Fig. 19] are:

(i) Casing: This is an annular channel of decreasing cross-sectional area about the axis, the tube

being of geometric shape of volute or a spiral. It is made either of cast steel or welded rolled steel

plates or concrete or concrete and steel. It completely surrounds the runner. The water first fills

the casing and then enters the guide vanes from all sides radially inwards. The decreasing cross-

sectional area helps the velocity of the entering water from all sides being equal. The volute or

spiral shape helps the entering water avoiding or preventing the creation of eddies.

The casing is also provided with inspection holes and pressure gauge connections.

(ii) Speed Ring or Stay Ring

From scroll casing, water passes through speed or stay rings. These rings are held together by

means of fixed vanes called Stay Vanes. The number of these stay vanes is generally half of the

number of guide vanes. Stay Rings serve two purposes:

(i) It directs the water from the scroll casing to the guide vanes.

(ii) It resists the load imposed on it by the internal pressure of water and also the weight

of turbine and the generator and transmits it to the foundation.

It is made either of cast iron or cast steel or fabricated steel.

Fig. 19 Francis Turbine

(iii) Guide Vanes

From speed ring, water passes through a series of guide vanes are called Wicket Gates. These

vanes are fixed between two rings in the form of a wheel known as guide wheel. Guide vanes

have two functions:

(i) to regulate the quantity of water supplied to the runner.

(ii) to direct the water on to the runner smoothly and with minimum loss of head i.e. at

appropriate design angle.

Each guide vane can rotate about its pivotal axis. The pivots of all vanes are connected to two

regulating rods through the regulation ring. The guide vanes are operated either by means of

hand wheel or automatically by governor.

(iv) Runner with vanes: The runner is mounted on a shaft and the blades are fixed on the runner

at equal distances. The vanes are so shaped that the water reacting with them will pass through

them thereby transferring their pressure energy to make it rotate the runner.

(iv) Draft tube: This is a divergent tube fixed at the end of the outlet of the turbine and the other

end is submerged under the water level in the tail race. The water after working on the turbine,

transfers the pressure energy there by losing all its pressure and falling below atmospheric

pressure. The draft tube accepts this water at the upper end and increases its pressure as the water

flows through the tube and increases more than atmospheric pressure before it reaches the

tailrace.

9.4 Design of Francis Turbine Runner

For a given value of power (kW) to be developed at had H when the turbine is to run at a speed R

r.p.m, following procedure is adopted to obtain preliminary dimensions:

(i). Assume suitable values of H

(85 to 90%) and o

(80 to 90%)

(ii) Obtain Q from )(1000

kWQH

P o

or ).(75

PHQH

o

(iii) Ratio of runner width to runner diameter is designated by n which varies from 0.1 to 0.45.

Assume a suitable value of n.

(iv) Ratio of flow velocity Vf1 at the inlet tip of the vane to spouting velocity gH2 is designated as Flow Ratio, .

gH

Vf

2

1 or gHVf

21

0.15 to 0.30

Chose appropriate value of and obtain Vf1.

(v) The area at inlet =111

kbD , where k1 accounts for space occupied by the thickness of runner

vanes. Assume k1 = 0.95.

1111111 )( knDD

Q

kbD

QV

f

Knowing Q, k1 and Vf1, D1 and b1 can be calculated.

(vi) The tangential velocity at the inlet u1 can be expressed in terms of spouting velocity through

speed ratio, i.e. gHu 21

, where 0.6 to 0.9.

Assume suitable value of and determine u1.

Also, 60

11

NDu

from where D1 can be determined and a check on D1 value determined in (v)

can be made.

(vii) Assume hydraulic efficiency value. For radial flow,gH

uVw

H

11 . Find Vw1.

(viii) Guide blade angle at inlet, , and vane angle at inlet, can be determined from:

1

1tan

w

f

V

V

and

11

1tanuV

V

w

f

(ix) Runner diameter D2 at the outlet 3

2

3

1to D1. Assume suitable D2 so that

60

22

NDu

.

(x) If t2 and b2 are respectively thickness of vane and the width of runner at the exit, then

2222 fVbDkQ Assume k2 = 0.95.

So that

111

222

1

2

bDk

bDk

V

V

f

f

Generally, it is assumed that 12 ff

VV and k1 = k2, so that

b2D2 = b1D1

(ix) For flow to be radial at exit of vanes, 0 and blade tip angle at the exit will be

2

1

2

2tanu

V

u

Vff

(xii) The number of guide blades, Zg, is kept even in order to avoid vibrations due to

synchronous effects. This number may vary from 8 to 24. Assume Zg.

Then number of runner blades, 1gr

ZZ . Thickness of runner blades ranges from 6mm to

24mm.

9.5 Kaplan Turbine

Fig. 20 Kaplan turbine installation

The reaction turbine developed by Victor Kaplan (1815-1892) is an improved version of the

older propeller turbine. It is particularly suitable for generating hydropower in locations where

large quantities of water are available under a relatively low head. Consequently the specific

speed of these turbines is high, viz., 300 to 1000. As in the case of a Francis turbine, the Kaplan

turbine is provided with a spiral casing, guide vane assembly and a draft tube. The blades of a

Kaplan turbine, three to eight in number are pivoted around the central hub or boss, thus

permitting adjustment of their orientation for changes in load and head. This arrangement is

generally carried out by the governor which also moves the guide vane suitably. For this reason,

while a fixed blade propeller turbine gives the best performance under the design load

conditions, a Kaplan turbine gives a consistently high efficiency over a larger range of heads,

discharges and loads. The facility for adjustment of blade angles ensures shock-less flow even

under non-design conditions of operation.

Water entering radially from the spiral casing is imparted a substantial whirl component by the

wicket gates. Subsequently, the curvature of the housing makes the flow become axial to some

extent and finally then relative flow as it enters the runner, is tangential to the leading edge of the

blade as shown in Fig 20(c), Energy transfer from fluid to runner depends essentially on the

extent to which the blade is capable of extinguishing the whirl component of fluid. In most

Kaplan runners as in Francis runners, water leaves the wheel axially with almost zero whirl or

tangential component. The velocity triangles shown in Fig 20(c) are at the inlet and outlet tips of

the runner vane at mid radius, i.e., midway between boss periphery and runner periphery.

Table below gives a comparison between Reaction and Impulse Turbines

S

N

Reaction turbine Impulse turbine

1 Only a fraction of the available hydraulic

energy is converted into kinetic energy

before the fluid enters the runner.

All the available hydraulic energy is

converted into kinetic energy by a nozzle

and it is the jet so produced which strikes

the runner blades.

2. Both pressure and velocity change as the

fluid passes through the runner. Pressure at

inlet is much higher than at the outlet.

It is the velocity of jet which changes, the

pressure throughout remaining atmospheric.

3 The runner must be enclosed within a Water-tight casing is not necessary. Casing

watertight casing (scroll casing). has no hydraulic function to perform. It only

serves to prevent splashing and guide water

to the tail race

4. Water is admitted over the entire

circumference of the runner

Water is admitted only in the form of jets. .

There may be one or more jets striking

equal number of buckets simultaneously.

5. Water completely fills at the passages

between the blades and while flowing

between inlet and outlet sections does

work on the blades

The turbine does not run full and air has a

free access to the buckets

6. The turbine is connected to the tail race

through a draft tube which is a gradually

expanding passage. It may be installed

above or below the tail race

The turbine is always installed above the

tail race and there is no draft tube used

7. The flow regulation is carried out by

means of a guide-vane assembly. Other

component parts are scroll casing, stay

ring, runner and the draft tube

Flow regulation is done by means of a

needle valve fitted into the nozzle.

10. Run Away Speed

When external load on the turbine drops to zero and if governing system also fails, then turbine

runner will race up and will attain maximum possible speed. This limiting speed of the runner is

called Run Away speed.

Pelton Turbine = 1.8 to 1.9 times normal speed; Francis Turbine = 2 to 2.2 times normal speed

Kaplan turbine = 2.5 to 3.0 times normal speed

Hence for safe design, the components of turbines are designed for runaway speed.

11. Unit Quantities

In the studies of comparison of the performances of turbines of different output, speeds and

different heads, it is convenient to determine the output, the speed and the discharge, when the

head on the turbine is reduced to unity, i.e. 1 m. The conditions of the turbine under unit head are

such that the efficiency of the turbine remains unaffected. Thus the velocity triangles under

working conditions and under unit head are geometrically similar.

Given a turbine every velocity vector (V1, U1, Vw1, Vf1) is a function of H where H is the head on

the turbine. With this basic concept, we can determine the speed, discharge and power under unit

head.

11.1 Unit Speed (Nu)

This is the speed of a turbine working under a unit head

Let N be the speed of turbine, H be the head on the turbine and u be the peripheral velocity.

We know that the peripheral velocity u is given by

60

NDu

where D is the mean diameter of the runner which is treated as constant and N is

the speed of the runner,

Hence u N

But HgKuu

2 and hence Hu

Hence HN

i.e. HKN1

, where K1 is the proportionality constant.

From definition of Unit speed, it is the speed of a turbine when working under unit head. Hence

at H=1, N=Nu. Substituting, we get

11

KNu (46)

HNuN or H

NNu (47)

11.2 Unit Discharge (Qu)

This is the discharge through the turbine working under a unit head.

Consider the Q as the discharge through a turbine. From discharge continuity equation,

Q = a x V, where a is the cross-sectional area of flow and V is the mean flow velocity.

For a given turbine, the cross-sectional area is constant and hence Q V

But HgCVv

2 and hence HV and hence HQ

i.e., HKQ2

where K2 is the proportionality constant.

From definition of Unit discharge, it is the discharge through the turbine when working under

unit head. Hence at H=1, Q=Qu. Substituting, we get

12

KQu (48)

HQQu

or H

QQ

u (49)

11.3 Unit Power (Pu)

This is the Power developed by the turbine working under a unit head.

Consider the P as the power developed by the turbine.

Efficiency of turbine, HQ

P

(50)

Where is the weight density of the fluid/water passing through the turbine, Q is the discharge

through the turbine and H is the head under which the turbine is working. But efficiency of a

turbine and weight density of water is constants and hence, we can write

P Q H (51)

From discharge continuity equation, Q = a x V, where a is the cross-sectional area of flow and V

is the mean flow velocity.

For a given turbine, the cross-sectional area is constant and hence Q V

But HgCVv

2 and hence HV and hence HQ

Substituting, we get

HHP or HHKP3

(52)

where K3 is the proportionality constant.

From definition of Unit Power, it is the power developed by the turbine when working under unit

head. Hence at H=1, P=Pu. Substituting, we get

113

KPu (53)

HHPuP or 2

3

H

P

HH

PPu (54)

Unit Speed, Unit discharge and Unit Power is definite characteristics of a turbine.

If for a given turbine under heads H1, H2, H3,. the corresponding speeds are N1, N2, N3,, the

corresponding discharges are Q1, Q2, Q3,. and the powers developed are P1, P2, P3,. Then

Unit speed =

3

3

2

2

1

1

H

N

H

N

H

NN

u

Unit Discharge =

3

3

2

2

1

1

H

Q

H

Q

H

QQ

u

Unit Power =2

3

3

3

23

2

2

23

1

1

3

3

2

2

1

1

H

P

H

P

H

PP or

HH

P

HH

P

HH

PP

uu

Thus if speed, discharge and power developed by a turbine under a certain head are known, the

corresponding quantities for any other head can be determined.

11.4 Specific Speed of a Turbine (Ns)

The specific speed of a turbine is the speed at which the turbine will run when developing unit

power under a unit head. This is the type characteristics of a turbine. For a set of geometrically

similar turbines the specific speed will have the same value.

Consider the P as the power developed by the turbine.

We know that the efficiency of turbine is given by

HQ

P

Where is the weight density of the fluid/water passing through the turbine, Q is the discharge

through the turbine and H is the head under which the turbine is working. But efficiency of a

turbine and weight density of water are constants and hence, we can write

Fig. 21 Pelton Turbine

P Q H

Discharge is given by the product of cross sectional area of flow and the flow velocity. Cross

sectional area is d b and hence

Q = D b Vf

But HgCVv

2 and hence HV

And HgKuu

2 and hence Hu

Further, the peripheral velocity is given by

60

NDu

where D is the mean diameter of the runner.

From the above two equations of u, we can write that

HDN or N

HD

But in turbines the width of flow area b is proportional to the diameter D. Hence D b, with

which

N

Hb

b

D

b

D

Hence HN

H

N

HP

or 2

25

N

HP

or P

HK

P

HN

25

25

2

Simplifying further

P

HKN

45

(55)

But from the definition of specific speed, it is the speed of a turbine when it is working under

unit head developing unit power. Hence when H = 1 and P = 1. N = Ns. Hence

ssNK andKN

1

1 45

(56)

Substituting we get

P

HNN

s

45

or 4

5

H

PNN

s (57)

12. Performance of Turbines

Turbines are usually designed fro particular values of head, speed, output and gate opening.

However, in actual practice they are to work under varying conditions. Hence, their behavioral

and their performance under these varying conditions are designed to be studied. For

convenience purpose, the test results are obtained from these types of studies are expressed in

unit quantities if comparison is required for the same types of turbines e.g. among Pelton

turbines. If the comparison is required between different types of turbines like Francis and Pelton

turbine for example, then the test results are expressed in terms of specific quantities.

(i) Performance under unit head

Consider a Francis turbine.

Fig. 22 Velocity Triangles for Francis turbine

The velocity triangles at the inlet when this turbine is subjected to the working head, H and unit

head will be same so that

uw

u

w V

u

V

u

1

1 (58)

uf

u

f V

u

V

u

1

1 (59)

From (58) and (59),

gV

guV

gV

guV

u

u

f

ww

f

w

2/

/

2/

/

22

1

1

1

1

1 (60)

Similarly for outlet

gV

guV

gV

guV

uu

u

f

uw

f

w

2/

/

2/

/22

22

2

(61)

This means that ratio of useful energy to wasted velocity energy is same under working head as

well as unit head. It means that the efficiency of the turbine at working head and unit head is

same i.e. uHH

or 1

11

g

uV

gH

uV uww u 1

1

gu

gHu

V

Vu

w

w

u

which leads to H

uuu

1 (62)

Equation 59 will lead to H

V

u

uVV

fu

ffu

1

1

1

(63)

Since HN

N

DN

DN

u

u

uuu

60/

60/1

, hence

H

NNu (64)

Equation 64 is called unit speed of the turbine.

Now HV

V

VDBk

VDBk

Q

Q

f

f

f

fuu u 1

)(

)(

11

or

H

QQu (65)

Equation 65 is called unit discharge.

Now P = ( QH) o and Pu = ( Qu1) ou = ( Qu) ou

Since o = ou for the same turbine, so

QH

Q

P

P uu = HH .

1 =

23

1

H Pu =

23

H

P (66)

Equation 66 is called unit power.

Once the performance under unit head is known for a turbine, then its possible to determine the

performance of that turbine under any other head, H1.

H

QQu =

1H

Q Q

H

HQ

21

11

(67)

23

H

PPu =

23

1

1

H

P P

H

HP

23

11

(68)

H

NNu =

1

1

H

N N

H

HN

21

11

(69)

(ii) Performance of turbine of the same type:

Here we are to define a unit turbine which is a turbine having the runner diameter of 1m and

operating under a head of 1m. Then

H

uuu

60

1 uN = H

ND 1

60

H

DNNu (70)

Similarly, HD

QQu 2

2

and 2

32 HD

PPu (71)

(iii) Performance of turbine of Different types

Here an imaginary turbine called SPECIFIC TURBINE is defined. This is a turbine which is

identical in shape, geometrical proportion, blade angle and gate opening etc. as actual turbine

but reduced to a size that will develop 1HP under unit head.

For actual turbine under unit head and for specific turbine under unit head, the inlet velocity

triangle will be same i.e. us = uu and Vfs = Vfu, so that

s

u

Q

Q=

fsss

fu

VnDD

VnDD

)(

)(

=

2

sD

D (72)

s

u

P

P =

oss

u

Q

Q

1

1 0 =

s

u

Q

Q=

2

sD

D (73)

Pu = Ps

2

sD

D=

2

sD

Das Ps = 1HP (74)

Ds =

uP

D Now Pu =

23

H

P Ds =

P

DH 23

(75)

Also uu = 60

uDN and us = 60

ss ND 60

uDN =60

ss ND

Ns = Nu

sD

D = Nu uP (76)

Also Nu = H

N and Pu =

23

H

P Ns =

45

H

PN (77)

Equation 77 is called SPECIFIC SPEED of the specific turbine or runner.

Similar relations can be developed for any other turbine.

Equation 77 also shows that specific speed is independent of dimensions and sizes of both

specific turbine and actual turbine. It means that All turbines of the same geometrical shape,

working under the same speed ratio, ku, and flow ratio, , and thus having the same efficiency,

will have the same specific speed, no matter what their size be and what power they develop

under what heads.

Hence Ns is the specific speed of the actual as well as specific turbine. Specific speed may be

defined as the speed of a turbine geometrically similar to the actual turbine but of such a size

that under corresponding conditions it will develop 1HP when working under unit head.

Specific speed is computed for operating conditions corresponding to maximum efficiency.

(See Books for this)

13. Characteristic curves of a Turbine

These are curves which are characteristic of a particular turbine which helps in studying the

performance of the turbine under various conditions. These curves pertaining to any turbine are

supplied by its manufacturers based on actual tests.

The data that must be obtained in testing a turbine are the following:

1. The speed of the turbine N

2. The discharge Q

3. The net head H

4. The power developed P

5. The overall efficiency o

6. Gate opening (this refers to the percentage of the inlet passages provided for water to

enter the turbine)

The characteristic curves obtained are the following:

1. Constant head curves or main characteristic curves

2. Constant speed curves or operating characteristic curves

3. Constant efficiency curves or Muschel curves

13.1 Constant head curves: Maintaining a constant head, the speed of the turbine is varied by

admitting different rates of flow by adjusting the percentage of gate opening. The power P

developed is measured mechanically. From each test the unit power Pu, the unit speed Nu, the

unit discharge Qu and the overall efficiency o are determined. The characteristic curves drawn

are

1. Unit discharge vs unit speed

2. Unit power vs unit speed

3. Overall efficiency vs unit speed

13.2 Constant speed curves: In this case tests are conducted at a constant speed varying the

head H and suitably adjusting the discharge Q. The power developed P is measured

mechanically. The overall efficiency is aimed at its maximum value.

The curves drawn are

P vs Q

o vs Q

o vs Pu

o max vs % Full load

13.3 Constant efficiency curves: These curves are plotted from data which can be obtained

from the constant head and constant speed curves. The object of obtaining this curve is to

determine the zone of constant efficiency so that we can always run the turbine with maximum

efficiency.

This curve also gives a good idea about the performance of the turbine at various efficiencies.

Fig. 23 Main Characteristic curves of a Pelton turbine

Fig. 24 Main Characteristic curves of a Kaplan turbine

Fig. 25 Main Characteristic curves of a Francis turbine

Fig. 26 Operating Characteristic curves of a turbine

Fig. 28 Constant efficiency curves for a reaction turbine

14. Draft Tubes

The water after working on the turbine, imparts its energy to the vanes and runner, thereby

reducing its pressure less than that of atmospheric pressure (Vacuum). As the water flows from

higher pressure to lower pressure, it cannot come out of the turbine and hence a divergent tube is

connected to the end of the turbine.

Draft tube is a divergent tube one end of which is connected to the outlet of the turbine and other

end is immersed well below the tailrace (Water level). The major function of the draft tube is to

increase the pressure from the inlet to outlet of the draft tube as it flows through it and hence

increase it more than atmospheric pressure. The other function is to safely discharge the water

that has worked on the turbine to tailrace.

Depending on the shape and alignment, draft tubes are classified as follows:

Straight Diverging

Hydracone

Simple Elbow

Elbow with change in shape

Hyperbolic tube

Fig. 27 Different types of Draft Tubes

Vertical divergent draft tube The draft tube has the shape of a frustum of a cone. This is

generally provided for low specific speed. The cone angle is not to exceed 8. For greater value

of the cone angle it is seen that the flowing body of water may not touch the sides of the draft

tube (Leaving the boundary). This will lead to the eddy formation bringing down the efficiency

of the draft tube.

Moodys draft tube or hydraucone: This is a bell mouthed draft tube or a conical tube with a

solid conical central core. The whirl of discharged water is very much reduced in this

arrangement.

Elbow draft tube: This draft tube affords to discharge the water horizontally to the tail race.

Elbow draft tube with circular inlet and rectangular outlet: This is a further improvement of the

simple elbow draft tube. In all the types mentioned above, the outlet of the draft tube should be

situated below the tail water level.

15. Model testing of turbines

In order to check the performance of the actual turbine, it is better to prepare a geometrically

similar model of the turbine. This model can tested under a known head, speed and flow rate and

its output i.e. power as well as the efficiency is determined. The model can be modified at lesser

cost of any modification required to meet the design.

The various variables involved are:

Discharge Q, head H, speed of rotation of runner N, runner diameter D, output power P, mass

destiny and viscosity .

),,,,,,(1 NDHgQfP (78)

However,

(1) It is convenient to use gH as one variable as it gives potential energy per unit mass as g as

individual parameters does not affect power.

(2) Since velocity is very high, Re, will be very large so that effect of viscosity can be ignored.

Hence, ),,,,(2 NDgHQfP (79)

Using dimensional analysis, one can write

3gHND

P

= f3

223

,DN

gH

ND

Q (80)

On the similar lines, one can write

= f4

223

,DN

gH

ND

Q (81)

In the above:

3ND

Q= Flow or Discharge number;

DN

gH2

= Head Number;

3gHND

P

= Power number

Eliminating D from Head number and power number, one gets:

21

45

21

)( gH

NP = Ns = Non-dimensional specific speed or Shape number.

For complete similarity to exist between the model and the actual (or prototype) turbines the

above noted parameters must have the same values for the model and the prototype turbines, i.e.

mND

Q

3

= pND

Q

3

or

mDgH

Q

2 =

pDgH

Q

2 (82)

mND

gH

2

= pND

gH

2

(83)

mgHND

P

3

=

pgHND

P

3

or

mDHg

P

223

23

=

pDHg

P

223

23

(84)

mDgH

PN

245

21

)( =

mDgH

PN

245

21

)( (85)

However, Reynolds number which should also be equal in the model and the prototype turbines

for complete similitude. But in actual practice, it is not possible to have the same value of the

Reynolds number in the model and the prototype turbines, on account of considerable difference

in their sizes, However, since the flow in the prototype turbine is turbulent, it has been observed

that if the flow in the model turbine is also turbulent, then even if the Reynolds numbers' are not

equal for the model and the prototype turbines the similarity between them can be ensured.

A little consideration will show that the above noted conditions may be achieved if, Ku, and

Ns have the same values for the model and the prototype turbines. Further it may be seen that the

various terms in the above expressions represent the unit quantities and hence for similarity

between the model and the prototype turbines the values of the various unit quantities must be

equal for the both.

If om and op are the overall efficiencies of the model and prototype, then

om = m

L

H

hH

and op =

p

L

H

hH

(86)

where H and hL are respectively the head acting on the turbine and the loss of head in the turbine.

So that the net effective heads for the model and the prototype turbines will be

(H hL)m = om (Hm) and (H hL)p = op (Hp) (87)

So that the above numbers in terms of efficiency can be written as:

moDH

Q

2 =

poDH

Q

2 (88)

m

o

DN

H

22

=

p

o

DN

H

22

(89)

moDH

P

223

=

poDH

P

223

(90)

moH

PN

45

45

=

poH

PN

45

45

(91)

It can be seen from the above that:

ps

ms

N

N

)(

)( =

45

op

om

(92)

For determining the efficiency of a prototype turbine from the efficiency obtained for its model,

the following general expression has been given:

om

op

1

1 =

p

m

D

D

p

m

H

H (93)

where om and op are the overall efficiencies of the model and the prototype respectively, Dm

and Dp are the respective diameters of their runners, Hm and Hp are the heads acting on the model

and the prototype turbines respectively and and are the numerical exponents. Different

values have been suggested for the exponents and by different investigators, but the most

commonly adopted values are those recommended by L.F. Moody which are = 0.20 and =

0.

16. Losses in Turbines

(i) Hydraulic Losses:

Hydraulic losses in the wheel are commonly expressed as a function of relative velocity at wheel

exit or

g

VKh

r

2

2

2 (94)

These are friction losses due to the movement of water through the runner passages and are,

therefore, for reaction turbine relatively larger for small wheels under high heads. It is due to this

reason that impulse turbines are more efficient than reaction turbine for high heads.

(ii) Disk Friction:

Denoted by hD, represents the power required to drive the runner submerged. The water above

and below the runner rotates at about half the speed of the runner, and hence, a friction loss

occurs between this water and the stationary part (the upper and lower covers) and an additional

friction loss occurs between this water and runner.

To minimize this loss, the inner surface of the covers and the outer surface of the wheel runner

must be made smooth and free from unnecessary projections. Disk friction is relatively greater

for high-head units, the power thus lost varying according to the formula, hD = KD5N

3, and is of

importance in affecting the efficiency of such units, especially at part gate.

(iii) Leakage Losses:

Denoted by hL occur due to escaping of water between the periphery of the runner and the

stationary parts, These are important with high head units and depend on the head and the area

between the runner seals and stationary parts.

These losses can be minimized by using rubber seal rings, which permit the operation of the

turbine with little or no clearance between the rotating and stationary parts and yet an elastic

surface well lubricated (by water) to accommodate inaccuracies of alignment without danger of

tearing or surging, or by the use of labyrinth seal for the runner.

Disk friction and leakage losses don't occur with impulse tu1'bine, although the windage losses

of the latter, due to air friction of the moving wheel and its parts, are somewhat analogous to disk

friction losses of the reaction wheel.

(iv) Draft Tube and Exit Losses

This includs friction as well as the velocity-head loss at exit from the tube. These may be

expressed as:

g

Vmh

r

d2

2

2 (95)

The value of m would be

2

3

2

A

A, which tend to make it somewhat larger and approach unity as

an extreme limit.

As no draft tube is used in impulse wheel; in this case m becomes unity, or in other words, the

entire velocity head at exit from the wheel is lost.

(v) Shock Losses:

Denoted by hs, occur in case the wheel is running at other than full-load point or at part gate.

These losses are caused by the direction of flow and its velocity at entrance to the wheel runner

being somewhat changed at part gate from the normal values for which the wheel runners are

designed.

This loss remains practically constant for the impulse turbine at all wheel gates; hence, its better-

sustained efficiency at part gate than the reaction wheel. So that the total effective head, h, is

given by:

sLD

r

u hg

Vmhh

g

VKhh

22

2

2

2

2 (96)

Not all of the head hu utilized by the wheel effectively reaches the generator as it gets reduced by

the mechanical losses, chiefly the friction of the wheel on its bearing and which vary with the

speed of the turbine in irregular manner between N and N2.

17. Cavitation in turbines

When the pressure in any part of the turbine reaches the vapour pressure of the flowing

water, it boils and small bubbles of vapour form in large numbers. These bubbles are carried

along by the flow, and on reaching the high pressure zones these bubbles suddenly

collapse as the vapour condenses to liquid again. The alternate formation and collapse of

vapour bubbles may cause severe damage to the surface which ultimately fails to fatigue and the

surface becomes badly scored and pitted. This phenomenon is known as cavitation.

In order to determine whether cavitation will occur in any portion of the turbine, D.

Thomas has developed a dimensionless parameter called Thomas' cavitation factor , , which is

expressed as

H

HHH sva (97)

Ha = atmospheric pressure head

Hv = vapour pressure head

Hs = suction pressure head

It has been found that depends on Ns.

For a turbine of particular Ns, can be reduced to such a value up to which o remains constant.

The limited value of at which o starts decreasing is called critical cavitation factor, c .

For Francis turbines: Critical cavitation factor; c

= 0.044 (Ns/100)2 (98)

For Kaplan turbine, c

= 1.1

3

100)0032.0(28.0 s

N (99)

It is, therefore, necessary that similarity w.r.t also exists between model and prototype to test the

performance of turbine.

18. Surge Tanks

When the load on the generator decreases the governor reduces the rate of flow of water striking

the runner in order to maintain the constant speed for the runner. But the sudden reduction of the

rate of flow in the penstock may lead to setting up of water hammer in the pipe, which may cause

excessive inertia pressure in the pipeline due to which the pipe may burst. Deflector and relief

valves are provided to 'avoid the sudden reduction of the rate of flow in the penstock.

But neither of these devices is of any assistance when the load on the generator increases and the

turbine is in need of more water. Thus in order to fulfill both the above noted requirements, in

addition to the above noted devices certain other devices such as surge tank and forebay are

usually employed.

Surge tanks are employed in the case of high and medium head hydro-power plants where the

penstock is very long, while forebays are suitable for medium and low head hydro-power plants

where the length of the penstock is short.

An ordinary surge tank is a cylindrical open-topped storage reservoir, as shown in Fig. 30, which

is connected to the penstock at a point as close as possible to the turbine.

Fig. 28 Surge Tanks and its types

The upper lip of tank is kept well above the maximum water level in the supply reservoir. When

the load on the turbine is steady and normal and there are no velocities variations in the pipeline

there will be a normal pressure gradient oaa1.

The water surface in the surge tank will be lower than the reservoir surface by an amount equal

to the friction head loss in the pipe connecting the reservoir and the surge tank. When the load on

the generator is reduced, turbine gates are closed and the water moving towards the turbine has

to move backward. The rejected water is then stored in the surge tank in the space between levels

o and b and a rising pressure gradient obbl is developed.

The retarding head so built up in the surge tank reduces the velocity of flow in the pipeline

corresponding to the reduced discharge required by the turbine.

When the load on the generator increases, the governor opens the turbine gates to increase the

rate of flow entering the runner. The increased demand of water by the turbine is partly met by

the water stored between levels a and c in the surge tank. As such the water level in the surge

tank falls and a falling pressure gradient occ1 is developed. In other words, the surge tank

develops an accelerating head which increases the velocity of flow in the pipeline to a value

corresponding to the increased discharge required by the turbine.

The various other types of surge tanks are also shown in Fig. 30.

Type (a) is a conical type surge tank,

Type (b) has an internal bell mouth spillway which permits the overflow to be easily disposed of.

Type (c) is known as the differential surge tank, which is provided with a central riser pipe

having small ports or holes at its lower end. The main advantage of this is that for the same

stabilising effects its capacity may be less than that of a simple cylindrical surge tank. This is so

because in a differential tank retarding and accelerating heads are developed more promptly than

in a simple surge tank in which the heads only built-up gradually as the tank fills. Moreover no

water is spilled to waste from the differential tank.

Type (d) is also similar in performance to the differential tank, but it is suitable when appropriate

.earth or rock excavation can be carried out.

19. Selection of turbines

(i) Based on Head and Specific speed

S. No. Head (m) Type of Turbine Specific Sped

1. 300 or more Pelton Wheel

Single or multiple jet

8.5 to 47 (SI units)

10-55 (metric)

2. 150-300 Pelton Wheel

Francis Turbine

30-85 (SI units)

35-100 (metric)

3. 60-150 Francis Turbine

Deriaz Turbine

85-188 (SI units)

100-220 (metric)

4. Less than 60 Kaplan turbine

Propeller turbine

Deriaz or tubular turbine

188-860 (SI units)

220-1000 (metric)

(ii) Part Load Operation: The variability of load will influence the choice of turbine if the head

lies between 150m to 300 m or lies below 30m. For higher ranges of heads, Pelton should be

used, for less than 30m Kaplan should be preferred.

(iii) Based on operational range

Fig. 29 Turbine selection based on operational range

Problems on Pelton or Impulse Turbines

Example

The head at the base of the nozzle of a Pelton wheel is 640 m. The outlet vane angle of the

bucket is 15o. The relative velocity at the outlet is reduced by 15% due to friction along the

vanes. If the discharge at outlet is without whirl find the ratio of bucket speed to the jet speed. If

the jet diameter is 100 mm while the wheel diameter is 1.2 m, find the speed of the turbine in

rpm, the force exerted by the jet on the wheel, the Power developed and the hydraulic efficiency.

Take Cv=0.97.

Solution:

H = 640 m; =15o; Vr1 = 0.85 Vr2; Vw2 = 0; d = 100 mm; D = 1.2 m;

Cv = 0.97; Ku = ?; N = ?; Fx = ?; P = ?; h = ?

We know that the absolute velocity of jet is given by

74.10964010297.02 HgCV v m/s

Fig. A Velocity triangle

Let the bucket speed be u

Relative velocity at inlet = Vr1 = V1-u = 109.74-u

Relative velocity at outlet = Vr2 = (1-0.15)Vr1 = 0.85(109.74-u)

But Vr2cos = u 0.85(109.74-u)cos15

Hence u = 49.48 m/s

u1 Vr1

V1=Vw1 u

Deflection angle

Vr2

u2 Vw2=0

V2=Vf2

Vf1=0

But 60

NDu

5.7872.1

48.496060

D

uN rpm (Ans)

Jet ratio = m = 45.074.109

48.49

V

u

Weight of water supplied = Q = 22

74.1091.04

100010

= 8.62kN/s

Force exerted = 211 wwx

VVVaF

But Vw1 = V1 and Vw2 = 0

58.9474.1091.04

100022

x

F kN

Work done/second = skNuFx /82.467948.4958.94

Kinetic Energy/second = 323

174.1091.0

41000

2

1

2

1

Va

= 5189.85kN/s

Hydraulic Efficiency = %17.9010085.5189

82.4679

Energy/s Kinetic

done/s Workh

Example

A PELTON wheel turbine is having a mean runner diameter of 1.0 m and is running at 1000

rpm. The net head is 100.0 m. If the side clearance is 20 and discharge is 0.1m3/s, find the

power available at the nozzle and hydraulic efficiency of the turbine.

Solution:

D = 1.0 m; N = 1000 rpm; H = 100.0 m; = 20o; Q = 0.1 m3/s;

WD/s = ? and h = ?

Assume Cv = 0.98

Fig. B Velocity triangle

We know that the velocity of the jet is given by

m/sHgCVv

83.43100010298.02

The absolute velocity of the vane is given by

m/sND

u 36.5260

10001

60

This situation is impracticable and hence the data has to be modified. Clearly state the

assumption as follows:

Assume H = 700 m (Because it is assumed that the typing and seeing error as 100 for 700)

Absolute velocity of the jet is given by

m/sHgCVv

96.11570010298.02

Power available at the nozzle is the given by work done per second

WD/second = Q H = g Q H = 1000x10x0.1x700 = 700 kW

Hydraulic Efficiency is given by

%07.96)20cos1(36.5296.11596.115

36.522cos1

2212

1

uVV

uh

Example

A Pelton wheel has a mean bucket speed of 10 m/s with a jet of water flowing at the rate of 700

lps under a head of 30 m. The buckets deflect the jet through an angle of 160. Calculate the

power given by water to the runner and the hydraulic efficiency of the turbine. Assume the

coefficient of nozzle as 0.98.

Solution:

u = 10 m/s; Q = 0.7 m3/s; = 180-160 = 20o; H = 30 m; Cv = 0.98;

WD/s = ? and h = ?

Assume g = 10m/s2

m/sHgCVv

243010298.02

Vr1 = V1-u = 24 10 = 14 m/s

Assuming no shock and frictional losses we have Vr1 = Vr2 = 14 m/s

Vw2 = Vr2 Cos - u = 14 x Cos 20 10 = 3.16 m/s

We know that the Work done by the jet on the vane is given by

WD/s 21211 wwww

VVuQuVVVa as Q = aV1

12.19016.324107.01000 kN-m/s (Ans)

Fig. C Velocity Triangle

Input per sec = KE/s 22

1

3

1247.01000

2

1

2

1

2

1 VQVa

201.6 kN/s

Hydraulic Efficiency = Output/Input = 190.12/201.6 = 94.305%

It can also be directly calculated by the derived equation as

94.29%uVV

uh

20cos1102424

102cos1

2212

1

(Ans)

Example

A Pelton wheel has to develop 13230 kW under a net head of 800 m while running at a speed of

600 rpm. If the coefficient of Jet Cy = 0.97, speed ratio = 0.46 and the ratio of the Jet diameter

is 1 /16 of wheel diameter. Calculate:

1. Pitch circle diameter

2. the diameter of jet

3. the quantity of water supplied to the wheel

4. the number of Jets required.

Assume overall efficiency as 85%.

Solution:

P = 13239 kW; H = 800 m; N = 600 rpm; Cv = 0.97; = 0.46 (Speed ratio)

m19.2

N

uD 22.1

300

6060

(Ans)

Single jet Pelton turbine is assumed.

The diameter of jet is given by the discharge continuity equation

171875.2.3944

220dVdQ

Hence d = 74.7 mm

The design parameters are

Single jet

Pitch Diameter = 1.22 m

Jet diameter = 74.7 mm

Jet Ratio = 32.160747.0

22.1

d

Dm

No. of Buckets = 0.5xm + 15 = 24

Example

It is desired to generate 1000 kW of power and survey reveals that 450 m of static head and a

minimum flow of 0.3m3/s are available. Comment whether the task can be accomplished by

installing a Pelton wheel run at 1000 rpm and having an overall efficiency of 80%.

Further, design the Pelton wheel assuming suitable data for coefficient of velocity and coefficient

of drag.

Solution:

P = 1000 kW; H = 450 m; Q = 0.3 m3/s; N = 1000 rpm; o = 0.8

Assume Cv = 0.98; Ku=0.45; = 1000 kg/m3; g = 10 m/s

2

74.04503.0100010

1010003

HQ

P

Input

Outputo

For the given conditions of P, Q and H, it is not possible to achieve the desired efficiency of

80%.

To decide whether the task can be accomplished by a Pelton turbine compute the specific speed

Ns.

45

H

PNN

s ;

Here N is the speed of runner, P is the power developed in kW and H is the head available at the

inlet.

3525.15

450

10001000

45

sN

Hence the installation of single jet Pelton wheel is justified.

Absolute velocity of jet is given by

m/sHgCVv

97.9245010298.02

Absolute velocity of vane is given by

19.2Hgu 8010248.02 m/s

The absolute velocity of vane is also given by

60

NDu

and hence

m19.2

N

uD 22.1

300

6060

(Ans)

Single jet Pelton turbine is assumed

The diameter of jet is given by the discharge continuity equation

171875.2.3944

220dVdQ

Hence d = 74.7 mm

The design parameters are

Single jet

Pitch Diameter = 1.22 m

Jet diameter = 74.7 mm

Jet Ratio = 32.160747.0

22.1

d

Dm

No. of Buckets = 0.5xm + 15 = 24

Example

A double jet Pelton wheel develops 895 MKW with an overall efficiency of 82% under a head of

60m. The speed ratio = 0.46, jet ratio = 12 and the nozzle coefficient = 0.97. Find the jet

diameter, wheel diameter and wheel speed in RPM.

Solution:

No. of jets = n = 2; P = 895 kW; o = 0.82; H = 60 m; Ku = 0.46; m = 12;

Cv = 0.97; D = ?; d = ?; N = ?

We know that the absolute velocity of jet is given by

6.336010297.02 HgCVv

m/s

The absolute velocity of vane is given by

93.156010246.02 HgKuu

m/s

Overall efficiency is given by

HQ

Po

and hence 819.16082.01010

108953

3

H

PQ

m

3/s

Discharge per jet = 9095.02

819.1

n

Qq m3/s

From discharge continuity equation, discharge per jet is also given by

md

dV

dq

186.0

9095.06.3344

22

Further, the jet ratio d

Dm 12

Hence D = 2.232 m

Also 60

NDu

and hence 136

232.2

93.156060

D

uN rpm

Note: Design a Pelton wheel: Width of bucket = 5d and depth of bucket is 1.2d

Example

The following data is related to a Pelton wheel:

Head at the base of the nozzle = 80m; Diameter of the jet = 100 mm; Discharge of the nozzle =

0.3m3/s; Power at the shaft = 206 kW; Power absorbed in mechanical resistance = 4.5 kW.

Determine

1. Power lost in the nozzle and

2. Power lost due to hydraulic resistance in the runner.

Solution

H = 80 m; d = 0.1m; a = d2 = 0.007854 m2; Q = 0.3 m3/s; SP = 206 kW; Power absorbed in

mechanical resistance = 4.5 kW.

From discharge continuity equation, we have,

Q = a x V = 0.007854 x V 0.3

V = 38.197 m/s

Power at the base of the nozzle = g Q H = 1000 x 10 x 0.3 x 80 = 240 kW

Power corresponding to the kinetic energy of the jet = a V3 = 218.85 kW

Power at the base of the nozzle = Power of the jet + Power lost in the nozzle

Power lost in the nozzle = 240 218.85 = 21.15 kW (Ans)

Power at the base of the nozzle = Power at the shaft + Power lost in the (nozzle +

runner + due to mechanical resistance)

Power lost in the runner = 240 (206 + 21.15 + 4.5) = 5.35 kW (Ans)

Example

The water available for a Pelton wheel is 4m3/s and the total head from reservoir to the

nozzle is 250 m. The turbine has two runners with two jets per runner. All the four jets

have the same diameters. The pipeline is 3000 m long. The efficiency of power

transmission through the pipeline and the nozzle is 91% and efficiency of each runner is

90%. The velocity coefficient of each nozzle is 0.975 and coefficient of friction 4f for

the pipe is 0.0045. Determine:

1. The power developed by the turbine;

2. The diameter of the jet

3. The diameter of the pipeline.

Solution

Q = 4 m3/s; Hg = 250 m;

No. of jets = n = 2 x 2 = 4;

Length of pipe = l = 3000 m;

Efficiency of the pipeline and the nozzle = 0.91

Efficiency of the runner = h = 0.9; Cv = 0.975; 4f = 0.0045

Efficiency of power transmission through pipelines and nozzle =

250

25091.0 f

g

fgh

H

hH

Hence hf = 22.5 m

Net head on the turbine = H = Hg hf = 227.5 m

Velocity of jet = m/s65.77HgCVv

5.227102975.021

(i) Power at inlet of the turbine = WP = Kinetic energy/second = a V3

WP = x 4 x 65.772 = 8651.39 kW

9.08651.39

turbine by developed Power

WP

turbine by developed Powerh

Hence power developed by turbine = 0.9 x 8651.39 = 7786.25 kW (Ans)

(ii) Discharge per jet = /smjets of No.

discharge Totalq

30.1

4

0.4

But 77.654

0.14

2

1

2 dVdq

Diameter of jet = d = 0.14 m (Ans)

(iii) If D is the diameter of the pipeline, then the head loss through the pipe is given by = hf

5

22

32

4

D

QLf

Dg

VLfh

f (From Q=aV)

5.223

430000045.05

2

D

hf

Hence D = 0.956 m (Ans)

Example

The three jet Pelton wheel is required to generate 10,000 kW under a net head of 400 m. The

blade at outlet is 15o and the reduction in the relative velocity while passing over the blade is 5%.

If the overall efficiency of the wheel is 80%, Cv = 0.98 and the speed ratio = 0.46, then find:

1. the diameter of the jet,

2. total flow

3. the force exerted by a jet on the buckets

4. the speed of the runner.

Solution:

No of jets = 3; Total Power P = 10,000 kW; Net head H = 400 m; Blade angle = = 15o; Vr2 =

0.95 Vr1; Overall efficiency = o = 0.8; Cv = 0.98; Speed ratio = Ku = 0.45; Frequency = f = 50

Hz/s.

We know that 400101000

10000,108.0

3

QHQg

Po

Q = 3.125 m3/s (Ans)

Discharge through one nozzle = /smn

Qq

3042.1

3

125.3

Velocity of the jet = /smHgCV3

v65.8740010298.02

1

But 65.874

042.14

2

1

2 dVdq

d = 123 mm (Ans)

Velocity of the Vane = /sm14.4140010246.02 3 HgKu u

Vr1 = (V1u1)=87.6541.14 = 46.51 m/s

Vr2 = 0.95 Vr1 = 0.95 x 46.51 = 44.18 m/s

Vw1 = V1 = 87.65 m/s

Vw2 = Vr2 cos u2 = 44.18 cos 1541.14 = 1.53 m/s

Force exerted by the jet on the buckets = Fx = q(Vw1+Vw2)

Fx = 1000 x 1.042 (87.65+1.53) = 92.926 kN (Ans)

Jet ratio = (Assumed)d

Dm 10

D = 1.23 m

60

NDu

Hence 23.1

14.416060

D

uN =638.8 rpm (Ans)

Problems on Reaction Tutbines

Example

The external diameter of an inward flow reaction turbine is 0.5 m. The width of the wheel at inlet

is 150 mm and the velocity of flow at inlet is 1.5 m/s. Find the rate of flow passing through the

turbine.

Solution:

D1 = 0.5 m, B1 = 0.15 m, Vf1 = 1.5 m/s, Q = ?

Discharge through the turbine = Q = D1 B1 Vf1 = x 0.5 x 0.15 x 1.5

Q = 0.353 m3/s (Ans)

Example

The external and internal diameters of an inward flow reaction turbine are 600 mm and 200 mm

respectively and the breadth at inlet is 150 mm. If the velocity of flow through the runner is

constant at 1.35m3/s, find the discharge through turbine and the width of wheel at outlet.

Solution:

D1 = 0.6 m, D2 = 0.2 m, B1 = 0.15 m, Vf1 = Vf2 = 1.35 m/s, Q = ?, B2 = ?

Discharge through the turbine = Q = D1 B1 Vf1 = x 0.6 x 0.15 x 1.35

Q = 0.382 m3/s (Ans)

Also discharge is given by Q = D2 B2 Vf2 = x 0.2 x B2 x 1.35 0.382

B2 = 0.45 m/s (Ans)

Example

An inward flow reaction turbine running at 500 rpm has an external diameter is 700 mm and a

width of 180 mm. If the guide vanes are at 20 to the wheel tangent and the absolute velocity of

water at inlet is 25 m/s, find (a) discharge through the turbine (b) inlet vane angle.

Solution:

N = 500 rpm, D1 = 0.7 m, B1 = 0.18 m, a = 20, V1 = 25 m/s, Q = ?, = ?

We know that the peripheral velocity is given by

smND

u /33.1860

5007.0

60

1

1

From inlet velocity triangle, we have

Vf1 = V1 Sin 20 = 8.55 m/s

Vw1 = V1 Cos 20 = 23.49 m/s

657.133.1849.23

55.8

11

1

uV

VTan

w

f

= 58.89 (Ans)

Q = D1 B1 Vf1 = x 0.7 x 0.18 x 8.55 = 3.384 m3/s (Ans)

Example

A reaction turbine works at 450 rpm under a head of 120 m. Its diameter at inlet is 1.2 m and the

flow area is 0.4 m2. The angle made by the absolute and relative velocities at inlet is 20 and 60

respectively with the tangential velocity. Determine (i) the discharge through the turbine (ii)

power developed (iii) efficiency. Assume radial discharge at outlet.

Solution:

N = 450 rpm, H = 120 m, D1 = 1.2 m, a1 = 0.4 m2, = 20 and = 60

Q = ?, = ?, Vw2 = 0

We know that the peripheral velocity is given by

smND

u /27.2860

4502.1

60

1

1

27.2860

1

1

11

1

w

f

w

f

V

VTan

uV

VTan

Hence Vf1 = (Vw1 28.27) Tan 60 (01)

Further 20

1

1 TanV

VTan

w

f

Hence Vf1 = (Vw1) tan 20 (02)

From equations 1 and 2, we get

(Vw1 28.27) tan 60 = Vw1 tan 20

Hence Vw1 = 35.79 m/s

Vf1 = 35.79 x tan 20 = 13.03 m/s

Discharge Q = D1 B1 Vf1 = a1 Vf1 = 0.4 x 13.03 = 5.212 m3/s (Ans)

Work done per unit weight of water =

NmkNuVg w

/178.10127.2879.3510

1111

Water Power or input per unit weight = H = 120 kN-m/N

Hydraulic efficiency = %31.84120

178.101

Example

The peripheral velocity at inlet of an outward flow reaction turbine is 12 m/s. The internal

diameter is 0.8 times the external diameter. The vanes are radial at entrance and the vane angle at

outlet is 20. The velocity of flow through the runner at inlet is 4 m/s. If the final discharge is

radial and the turbine is situated 1 m below tail water level, determine:

The guide blade angle

The absolute velocity of water leaving the guides

The head on the turbine

The hydraulic efficiency

Solution:

u1 = 12 m/s, D1 = 0.8 D2, = 90, = 20 , Vf1 = 4 m/s,Vw2 = 0, Pressure head at outlet = 1m, =

?, V1 = ?, H = ?, h = ?

From inlet velocity triangle,12

4

1

1 u

VTan f , Hence = 18.44

Absolute velocity of water leaving guide vanes is

m/s VuVf

65.12412222

1

2

11

6060

2

2

1

1

NDu and

NDu

Comparing the above 2 equations, we have

2

2

1

1

2

2

1

16060

D

u

D

u hence and

D

u

D

u

Hence m/suD

Du 15

8.0

12

1 12

2

From outlet velocity triangle, V2 = Vf2 = u2 tan 20 = 15 tan 20 = 5.46 m/s

As Vw2 = 0

Work done per unit weight of water = m/NkNg

uVw

4.14

10

121211

Head on turbine H

Energy Head at outlet = WD per unit weight + losses

g

uVw

g

VH 11

2

2

21 and hence

mg

H 89.164.142

46.51

2

Hydraulic efficiency = %26.8510089.1610

121211

Hg

uVw

h

Example

An inward flow water turbine has blades the inner and outer radii of which are 300 mm and 50

mm respectively. Water enters the blades at the outer periphery with a velocity of 45 m/s making

an angle of 25 with the tangent to the wheel at the inlet tip. Water leaves the blade with a flow

velocity of 8 m/s. If the blade angles at inlet and outlet are 35 and 25 respectively, determine

1. Speed of the turbine wheel

2. Work done per N of water

Solution:

D1 = 0.6 m; D2 = 0.1 m, V1 = 45 m/s, = 25, V2 = 8 m/s, = 35, = 25, N = ?, WD/N = ?

423.025

1

1 SinV

VSin f

Hence Vf1 = 0.423 x 45 = 19.035 m/s

466.025tan

1

1

w

f

V

VTan

Hence Vw1 = 40.848 m/s

111

1

848.40

035.107.035tan

uuV

VTan