types of tubomachines

1. Turbomachinery

2. • Rotating machine that adds or extracts energy from a
fluid by virtue of a rotating system of blades
• Hydraulic machines can be divided into displacement
machines and rotodynamic machines
• In displacement machines the volume of a chamber is
increased/decreased by forcing a fluid into and out of the
chamber. e.g. tyre pump, human heart etc.

3. • Rotodynamic machines have a set of blades, buckets,
flow channels/ passages forming a rotor. Its rotation
produces dynamic effects to extract/add energy from/to a
fluid.
• Includes turbines and pumps
• Have rotating element through which the fluid passes
• The rotor is called a runner in turbine and an impeller in
the pump.

4. • Turbine is a device that extracts energy from a fluid
(converts the energy held by the fluid to mechanical
energy)
• Pumps are devices that add energy to the fluid (e.g.
pumps, fans, blowers and compressors).

5. Turbines
• Hydro electric power is the most remarkable
development pertaining to the exploitation of
water resources throughout the world
• Hydroelectric power is developed by hydraulic
turbines which are hydraulic machines.
• Turbines convert hydraulic energy or hydro-
potential into mechanical energy.

6. • Mechanical energy developed by turbines
is used to run electric generators coupled
to the shaft of turbines
• Hydro electric power is the most cheapest
source of power generation.

7. • Poncelet first introduced the idea of the
development of mechanical energy
through hydraulic energy
• Modern hydraulic turbines have been
developed by Pelton (impulse), Francis
(reaction) and Kaplan (propeller)

9. Classification of Turbines
• On the basis of hydraulic action or type of energy at the inlet
– Impulse Turbine (Pelton wheel)
– Reaction Turbine (Francis turbine and Kaplan)
• On the basis of direction of flow through the runner
– Tangential flow turbine (pelton)
– Radial flow turbine (francis )
– Axial Flow Turbine (Kaplan)
– Mixed flow turbine (modern francis)

10. • On the basis of head of water
– High head turbine (pelton, H>250m)
– Medium head turbine (modern francis, 45-250m)
– Low head turbine (kaplan, <45m)
• On the basis of specific speed, Ns, of the turbine
– Low specific speed (pelton, 10-35)
– Medium (francis, 60-400)
– High specific speed (kaplan, 300-1000)
Specific speed is the speed of turbine for producing unit power under
unit head

11. • On the basis of head of water
– High head turbine (pelton, H>250m)
– Medium head turbine (modern francis, 45-250m)
– Low head turbine (kaplan, <45m)
• On the basis of specific speed, Ns, of the turbine
– Low specific speed (pelton, 10-35)
– Medium (francis, 60-400)
– High specific speed (kaplan, 300-1000)
Specific speed is the speed of turbine for producing unit power under
unit head

12. Energy Transfer
Basic equation of energy transfer in
rotodynamic machines

13. • The basic equation of fluid dynamics relating to
energy transfer is same for all rotodynamic
machines and is a simple form of " Newton 's
Laws of Motion" applied to a fluid element
traversing a rotor.
• Here we shall make use of the momentum
theorem as applicable to a fluid element while
flowing through fixed and moving vanes.
• Figure represents diagrammatically a rotor of a
generalised fluid machine, with 0-0 the axis of
rotation and ω the angular velocity.
• Fluid enters the rotor at 1, passes through the
rotor by any path and is discharged at 2.

14. • The points 1 and 2 are at r1 radii and r2 from the
centre of the rotor, and the directions of fluid
velocities at 1 and 2 may be at any arbitrary
angles. For the analysis of energy transfer due
to fluid flow in this situation, we assume the
following:
• (a) The flow is steady, that is, the mass flow
rate is constant across any section (no storage
or depletion of fluid mass in the rotor).
• (b) The heat and work interactions between the
rotor and its surroundings take place at a
constant rate.

15. • (c) Velocity is uniform over any area normal to
the flow. This means that the velocity vector at
any point is representative of the total flow over
a finite area. This condition also implies that
there is no leakage loss and the entire fluid is
undergoing the same process.

16. • The velocity at any point may be resolved into
three mutually perpendicular components as
shown in Fig. The axial component of velocity
Va is directed parallel to the axis of rotation , Vf
the radial component is directed radially through
the axis to rotation, while the tangential
component Vw is directed at right angles to the
radial direction and along the tangent to the rotor
at that part.
• The change in magnitude of the axial velocity
components through the rotor causes a change
in the axial momentum. This change gives rise
to an axial force, which must be taken by a
thrust bearing to the stationary rotor casing.

17. • The change in magnitude of radial velocity
causes a change in momentum in radial
direction.
• However, for an axisymmetric flow, this does not
result in any net radial force on the rotor. In case
of a non uniform flow distribution over the
periphery of the rotor in practice, a change in
momentum in radial direction may result in a net
radial force which is carried as a journal load.
The tangential component only has an effect on
the angular motion of the rotor.
• No torque is produced by axial and radial
components.

18. • Torque is exerted on the rotor only due to the
change in momentum of the tangential
component
• At inlet moment of momentum /mass
Vw1 r1
• At outlet moment of momentum / mass
Vw2 r2
Rate of change of moment of momentum =
m (Vw1 r1 - Vw2 r2)
m (Vw1 r1 – Vw2 r2) = T (Angular momentum theorem)

19. Rate of Energy imparted
E = T ω
E = m (Vw1 r1 – Vw2 r2) ω
E = m (Vw1 r1 ω – Vw2 r2 ω)
Where
r1 ω = Tangential / linear velocity at inlet = u1
r2 ω = Tangential / linear velocity at outlet = u2
E = m (Vw1 u1 - Vw2 u2) (1)
E/m = (Vw1 u1 - Vw2 u2) (2)

20. Equation 1, 2 and 3 are different form of a single
equation which is known as Euler’s equation.

21. Pelton Wheel Turbine
• Most commonly used impulse or tangential flow
turbine
• Named after its pioneer Leston A Pelton (1829-
1908).
• Suitable to be used for high head hydroelectric
power plants

22. Pelton Wheel Turbine
• In impulse turbine (Pelton wheel), the water from
a dam is made to flow through a pipeline, and
then the guide mechanism and finally through
the nozzle.
• In such a process, the entire available energy of
the water is converted into kinetic energy, by
passing it through the nozzles; which are kept
close to the runner.
• The water enters the running wheel in the form
of a jet ( or jets), which impinges on the buckets,
fixed to the outer periphery of the wheel.

23. Pelton Wheel Turbine
• The jet of water impinges on the buckets with a
high velocity, and after flowing over the vanes,
leaves with a low velocity; thus imparting energy
to the runner.
• The pressure of the water, both at entering and
leaving the vanes, is atmospheric.

24. Components
The Pelton wheel has the following components:
1.Nozzle
2. Runner and Buckets
3. Casing and
4. Breaking jet

26. 1. Nozzle with guide mechanism
• Function is to convert pressure energy to high velocity
energy in the form of jet.
• A spear is provided in the nozzle to control the flow due
to varying load on the turbine.
• Nozzle is made of either cast iron or cast steel
• Nozzle mouth ring and spear tip are made of non-
abrasive material (stainless steel or bronze) and can
easily be replaced
• Sudden closure of nozzle(s) results in sudden increase
in pressure which may burst the pipe; in order to avoid
such mishap, an additional nozzle ( known as bypass
nozzle) is provided through which the water can pass,
without striking the buckets.

27. • Sometimes a plate (known as deflector) is provided to
the nozzle, which is used to deflect the water jet, and
preventing it from striking the buckets.
• The nozzle is kept very close to the buckets, in order to
avoid minimise the losses due to windage.

28. 2. Runner and Buckets
• Runner is a circular disc with a number of evenly
spaced vanes or buckets semi-ellipsoidal in
shape
• Each bucket is divided into two symmetrical
compartments by a sharp edge ridge called
splitter
• Jet of water normally impinges on the splitter
dividing into two parts and leaving at the outer
edge

29. • To get the full reaction of the jet, it has to
be turned through 180 degrees but it may
strike the incoming bucket thus retarding
its speed.
• The angle through which the jet is turned
is normally kept between 160 and 170
degrees.

30. • As the splitter takes the full impact of the jet, so
it has to be quite strong and should not be
having a sharp edge
• To avoid erosion of buckets due to impurities
present in water, cast iron buckets are used for
low head plants while cast steel, stainless steel
and bronze are used for medium head plants
• Buckets are either cast as an integral part or are
bolted to the rim.

32. 3. Casing
• It does not have any hydraulic function
• Provided to avoid accidents, splashing of
water and to lead the water to the tail race.
• Made in two parts to facilitate assembling
• Material used is usually cast iron.

33. 4. Breaking Jet
• Whenever the turbine has to be brought to
rest, the nozzle is completely closed. It
has been observed that it goes on
revolving for a considerable time, due to
inertia, before it comes to rest.
• In order to bring the runner to rest in a
short time, a small nozzle is provided in
such a way that it will direct a jet of water
on the back of the buckets. It acts as a
break for reducing the speed of the

34. Work Done by an Impulse
Turbine
Inlet triangle

35. Outlet triangle

36. Let
V = Absolute velocity of the entering water
Vr = Relative velocity of water and bucket at inlet
Vf = Velocity of flow at inlet
V1, Vr1, Vf1 = Corresponding values at outlet
D = Diameter of the wheel
d = Diameter of the nozzle
H = Total head of water under which wheel is
working
= Angle of the blade tip at outletφ

37. Inlet velocity triangle is straight line as shown in
Fig.
From velocity triangle at inlet
As a matter of fact, the shape of the outlet velocity
triangle depends upon the value of Vw1. If is in the
same direction as that of jet, its value is taken as
positive. However, if is in the opposite direction (
as shown in the fig. ) its value is taken
vVVr
Vf
VwV
−=
=
=
0

38. as negative. The relationship between these two
velocity triangles is
Work done per kN of water
)(1
1
vVVV
vvv
rr −==
==
( )11/ vVvVmE ww −=
( )111 cos vVV rw −= φ
( )11/1/ vVvVgmgE ww −=
( )11)(/1/ vVvVgmgE ww −−=

39. ( )vVVgmgE ww )(/1/ 1+=
( ))cos(// 11 vVVgvmgE rw −+= φ
( )vVvVgmgE ww 1(/1/ +=
( ))cos((// 1 vVVgvmgE r −+= φ
( ))cos(// vVvVgvmgE rr −++= φ
( ))cos1(// φ+= rVgvmgE
( )φcos1)(// +−= vVgvmgE

40. Hydraulic Efficiency
= Work done per kN of water / Energy supplied per
kN of water
2
2
1
)cos1)((/
V
g
vVgv
h
φ
η
+−
=
( )( )
2
cos12
V
vvV φ+−
=

41. For maximum efficiency
0=
dv
d hη
( ) ( )
2/
042
022
cos1 2
2
Vv
vV
vVv
dv
d
V
=
=−⇒
=−
+ φ

42. • It means that the velocity of the wheel, for
maximum hydraulic efficiency, should be
half of the jet velocity. Therefore,
maximum work done / kN of water
( )φcos1)(/ +−= vVgv
( )φcos1)(2/2/ +−= VVgV
( )( )gV 4/cos1 2
φ+=

43. • Maximum hydraulic efficiency
( )
2
cos1 φ
η
+
=h
( )( )
)2/(
4/cos1
2
2
gV
gVφ+
=

44. Power Produced by an Impulse
Turbine
• Some work is done per kN of water, when
the jet strikes the buckets of an impulse
turbine. If we know the quantity of water
flowing through jet per second and the
amount of work done per second, the
power produced can be calculated as
(kW)
w = Specific weight of water
H = Head of water
Q = Discharge
wQHP =

45. Efficiencies of an Impulse
Turbine
• In general the term efficiency may be
defined as the ratio of work done to the
energy supplied.
• An impulse turbine has the following three
types of efficiencies
1.Hydraulic efficiency
It is the ratio of work done, on the wheel, to
the energy of the jet
Already calculated

46. Efficiencies of an Impulse
Turbine
2. Mechanical efficiency
It has been observed that all the energy
supplied to the wheel does not come out
as useful work. But a part of it is
dissipated in overcoming friction of
bearings and other moving parts. Thus the
mechanical efficiency is the ratio of actual
work available at turbine to the energy
imparted to the wheel.

47. Efficiencies of an Impulse
Turbine
3. Overall efficiency
It is the measure of the performance of a
turbine and is the ratio of actual power
produced by the turbine to input energy to
the turbine.
wQH
P
o =η

48. Number of Jets of a Pelton
Wheel
• A Pelton turbine, generally, has a single
jet only. But whenever a single jet can not
develop the required power, we may have
to employ more than one jets.
• While designing the jets care should
always be taken to provide the jets are
equidistant on the outer periphery of the
wheel.

49. Number of Jets of a Pelton
Wheel

50. Design or working proportions
of pelton wheel
A Pelton wheel is designed to find out the following data:
1. Diameter of the wheel
2. Diameter of the jet
3. Size ( i.e. Width and depth ) of the buckets
4. No. of Buckets
While designing a Pelton wheel, if sufficient data is not
available then the following assumptions are made,
which are meant for the best results:
1. Overall efficiency between 80% and 87% ( preferably
85%)
2. Coefficient of velocity 0.99 (preferably 0.985)

51. Design or working proportions
of pelton wheel
3. Ratio of peripheral velocity to the jet velocity as 0.46

52. Dimensions of bucket
• Width B of the bucket is normally taken as 4 to 5 d (d =
diameter of the jet)
• The depth of the bucket (c) normally 1.2 d
• Length L of the bucket is 2.4 to 3.2d
• Other dimensions are
= 10 to 15 degreesɸ
β1 = 5-8 degrees

53. Number of buckets
• The number of buckets is decided such that the
frictional loss is minimum and the path of the jet
is not disturbed.
• Also the jet must be fully utilised
• Taygun gave the following relation for the
calculation of number of buckets.
D = Mean Bucket diameter
d = Diameter of the jet
155.015
2
1
+=+=
d
D
d
D
bn

54. Design or working proportions
of pelton wheel
If a turbine is working under a net head H, then the ideal
velocity of the wheel is given by
But due to the frictional loss, the actual velocity is slightly
less than this, so the velocity V of jet at inlet
Cv (coefficient of velocity ranges from 0.97 to 0.99)
gH2
gHCV v 2=

55. Although, theoretically,
But actually, occurs when
v= 0.46V
If v is expressed in terms of speed ratio (ratio of tangential
velocity of wheel to theoretical velocity of jet), the speed
ratio of a pelton turbine is given by
Ku ranges from 0.43 to 0.47
2/Vv =
maxhη
gH
v
Ku
2
= gHv 246.0=

56. • The angle through which the jet is deflected is taken as
165 degree and β at the outlet velocity triangle is 15
degrees.
• Least diameter of the jet is given by
2/1
2/1
2/1
2
2
5416.0
)
298.0
4
(
2
4
2
4






=






××
=
=
==
H
Q
H
Q
g
d
HgC
Q
d
gHCdaVQ
v
v
π
π
π

57. If D is the mean diameter of the wheel and
N is the rotation of the wheel in rpm
N
gHK
N
v
D
DN
v
u
ππ
π
2.6060
60
==
=

58. Ratio of diameter of the runner to the least
diameter of the jet is known as jet ratio D/d
D/d is taken between 11 to 14 for maximum
efficiency
Normal value is taken as 12 if not given

59. 1. A Pelton wheel develops 2000 kW under a head of 100 m, and
with an overall efficiency of 85%. Find the diameter of the nozzle, if
the coefficient of velocity for the nozzle is 0.98.
2. A Pelton wheel having semi-circular buckets and working under a
head of 140 m is running 600 rpm. The discharge through the
nozzle is 500 lit/s and diameter of the wheel is 600 mm. Find
a) Power available at the nozzle
b) Hydraulic efficiency of the wheel, if coefficient of velocity is 0.98)
3. A Pelton wheel, working under a head of 500 m, produces 13000
kW at 430 rpm. If the efficiency of the wheel is 85%. Determine (a)
discharge of the turbine (b) diameter of the wheel and (c) diameter
of the nozzle. Assume suitable data.

60. 4. In hydraulic scheme, the distance between high level reservoir at the
top of mountains and turbine is 1.6 km and difference of their
levels is 500 m. The water is brought in 4 penstocks each of 0.9 m
connected to a nozzle of 200 mm diameter at the end. Find:
(a) Power of each jet and
(b) total power available at the reservoir, taking the value of
Darcy’s coefficient of friction as 0.008
5. A power house is equipped with impulse turbines of pelton type.
Each turbine delivers a maximum power of 14250 kW, when
working under a head of 900 m and running at 600 rpm.
Find the diameter of the jet, and the mean diameter of the wheel.
Take overall efficiency of the turbine as 89.2%.

61. 6. A Pelton wheel is required to generate 3750 kW under an effective
head of 400 m. Find the total flow in lit/s and size of the jet.
Assume generator efficiency 95%, overall efficiency 80%,
coefficient of velocity 0.97, speed ratio 0.46. If the jet ratio is 10,
find mean diameter of runner.
7. A Pelton wheel has a mean bucket speed of 15m/s with a jet of
water impinging with a velocity of 40m/s and discharging 450lit/s. If
the buckets deflect the jet through an angle of 165o
, find the power
generated by the wheel.
8. A Pelton wheel has a tangential velocity of buckets of 15 m/s. The
water is being supplied under a head of 150 m at the rate of 200
lits/s. The buckets deflects the jet through an angle of 160. If the
coefficient of velocity for the nozzle is 0.98, find the power
produced by the wheel and its hydraulic efficiency.

62. 9. A Pelton wheel is supplied water under a head of 200 m through a
100 mm diameter pipes. If the quantity of water supplied to the
wheel is 1.25 cumecs. Find the number of jets. Assume coefficient
of velocity 0.97.
10. A Pelton wheel has to develop 5000 kW under a net head of 300 m,
while running at a speed of 500 rpm. If the coefficient of velocity for
the jet = 0.97, speed ratio = 0.46 and the ratio of the jet diameter is
1/10 of wheel diameter, calculate (a) quantity of water supplied to
the wheel (b) diameter of pitch circle (c) diameter of jets and (d)
number of jets.
11. Design a Pelton wheel for a head of 350 m at a speed of 300 rpm.
Take overall efficiency of the wheel as 85% and ratio of the jet to
the wheel diameter as 1/10.

63. 12. Design a Pelton wheel for the for following data:
Head of water = 150 m
Power to be developed = 600 kW
Speed of wheel = 360rpm
Assume, reasonably, the missing data.