This document discusses Pelton wheel turbines. It begins with an overview of Pelton wheels and their components. It then provides explanations of key concepts such as impulse turbines, velocity diagrams, effective head, maximum power output, and hydraulic efficiency. Practical considerations for Pelton wheel design like optimal bucket angles are also covered. Finally, it discusses turbine selection and the typical range of specific speeds for different turbine types.

1. Minia University
‫ـــعة‬‫م‬‫جا‬
‫يا‬‫ن‬‫مل‬‫ا‬
CVE 313 Hydraulics II “Civil Engineering Department”

3. Pelton Wheel Turbine
By: D. Ezzat El sayed G. SALEH

5. 5
Hydrologic Cycle

8. Section through a Shaft Spillway
Morning Glory Spillway
with Anti-vortex Piers

9. What is Involved?
What steps are taken when constructing a Power
station?
• Preparing the Site
• Setting Foundations
• Building the Framework
• Enclosing the Structure
• Installing Utilities
• Finishing the Exterior and Interior
• Completing the Site

10. 10
Cover Dam

11. The repair of the damaged parts of the water-management apron and improvement
of the bottom behind the apron.

14. sheet pile wall

16. Pelton Wheel

19. Pelton
Wheel

21. Old Pelton wheel from , Walchensee Power Plant (Germany)

22. Assembly of a Pelton wheel
at Walchensee Power Plant

25. Head range
Turbine Type
50 > H <1300
Pelton Wheel
10 > H < 350
Francis Turbine
2 > H < 40
Kaplan Turbine
Typical Range of Heads

27. Reaction Turbine

28. A conventional dammed-hydro facility (hydroelectric dam) is the
most common type of hydroelectric power generation.

31. The runner of
the small water
turbine

32. A Francis Turbine runner, rated at nearly one million hp (750 MW),
being installed at the Grand Coulee Dam, United States.

33. Turbine row at Los Nihuiles Power Station (Mendoza, Argentina)

34. The Hoover Dam in the United States is a large
conventional dammed-hydro facility, with an
installed capacity of 2,080 MW .

35. The Three Gorges Dam is the largest operating hydroelectric power station,
at 22,500 MW.

36. A Francis Turbine at the end of its life showing cavitation pitting, fating
cracking and a catastrophic failure. Earlier repair jobs that used stainless steel
weld rods are visible.

37. Kaplan Turbine

38. Kaplan Turbine and Electric Generator
cut-away view

40. The Pelton Wheel Turbine
Definition
Out put Shaft Power,
Hydraulic Efficiency,
Velocity Diagram of Flow into and Out of the
Buckets,
 Effective Head for Pelton Wheel

41. *The high speed water coming out of the nozzle strikes
the splitter which divides the jet into two equal
streams. These stream flow along the inner curve of the
bucket and leave it in the direction opposite to that of
incoming jet. The high pressure water can be obtained
from any water body situated at some height or streams
of water flowing down the hills.
*The change in momentum (direction as well as speed)
of water stream produces an impulse on the blades of
the wheel of Pelton Turbine. This impulse generates the
torque and rotation in the shaft of Pelton Turbine.
41

42. *Nozzle: It controls the amount of water striking the
vanes of the runner.
*Casing: It is used to prevent splashing of water and
plays no part in power generation.
*Runner with buckets: Runner is a circular disc on
the periphery of which a number of evenly spaced
buckets are fixed.
*Breaking Jet: To stop the runner in short time
breaking jet is used.

44. *Runner of a Pelton Turbine
SPLITTER
BUCKETS OR VANES
RUNNER
44

45. Impulse turbines change the velocity of a water
jet. The jet pushes on the turbine's curved blades
which changes the direction of the flow.
The resulting change in momentum (impulse(
causes a force on the turbine blades. Since the
turbine is spinning, the force acts through a
distance (work) and the diverted water flow is left
with diminished energy

46. 
1800- 
Q
Q/2
Q/2
- Fb
nozzle
Q = AJ . VJ
Where:
AJ = the cross section area of the
incoming jet

47. 
1800- 
V
j
Vr (2) = k Vr
(1)
Vb
Vr (1) = Vj -
Vb
k Vr (1)



48. 9/21/2022 48
Applying the momentum
equation in the x-direction,
-Fb = -2 (  Q/2  V2 cos (180 -  )) -  Q V1 , or
Fb =  Q (Vr (1) + Vr (2) cos (180 -  ))
D
m

1800- 
Vj
Vr (2) = k Vr (1)
Vb
Vr (1) = Vj - Vb
k Vr (1)


Subscripts (1) and (2) refers to the relative jet velocity entering and
exiting the control volume, respectively.
Force

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Fx =  Q Vr (1) (1 + k cos (180 -  ))
Since Vr (2) = k  Vr (1)
The active force exerted on the bucket by the jet
Fb =  Q ( Vj – Vb ) (1+ k cos (180 -  ))
Where (Vj – Vb) is the velocity of the jet relative to the bucket
For a given (Vj – Vb), Fx will be maximized if  = 180o . Then the
above equation gives,
Fb = 2  Q ( Vj – Vb ) (k = 1.0)
 = 180o


50. 9/21/2022 50
 The torque on the shaft is given by, T = Fx  Dm /2
T =  Q ( Vj – Vb ) (1+ k cos (180 -  ))  (Dm /2)
Where Dm is the mean diameter of the wheel
 The power developed will be the product of the of the
force acting on the bucket and its speed

Torque
Power

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The flow rate (discharge) Q = Aj  Vj
P = F  Vb =F  (  Dm /2) =T  
=  Q  (Dm /2)  ( Vj – Vb ) (1+ k cos (180 -  ))
Where:
Aj = the cross section area of the incoming jet,
and
Vj = the velocity of the incoming jet.
Flow rate (Discharge)



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The corresponding hydraulic efficiency,
h = P / ( g Q H) = P / ( Q Vj
2/2 )
= 2 (Vb / Vj ) ( 1 – Vb / Vj ) (1+ k cos (180 -  ))
P =  Aj Vj
3  (Vb / Vj ) ( 1 – Vb / Vj ) (1+ k cos (180 -  ))
Substituting for Q , the power equation can be transformed
into,

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Maximum Power

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 If Vb = 0 or if Vb = Vj the power will be zero. There must
be an optimum bucket speed between there two limits
that will maximize the power. This speed can be
determined by differentiating P with respect to Vb and
setting the differential equal to zero,
P =  Aj Vj
3  (Vb /Vj ) ( 1 – Vb /Vj ) (1+ k cos (180 -  ))
(Vb /Vj ) = 0.5
 Thus, the maximum power will be developed when the
bucket speed is one half the jet speed.

55.  = Vb /Vj
1.0
0.5
Ideal Power
Actual Power
P/  Aj Vj
3
(1+
k
cos
(180-))/
2
 = Vb / Vj is between 0.46 and 0.48
 is the speed ratio

56. Net Head or
Effective Head for
Pelton Wheel

57. Vj
Vp

H
gross
H
net
H
L
Hnet - Hgross - HL
Hnet =Hgross – f L Vp
2 / (2 g Dp )
Thus,
Vj / VP = (Dp / Dj )2 Continuity Equation
Vj
2 /2g = Hnet
Penstock
Reservoir

58. For a perfect nozzle, the entire available head
would be converted to jet velocity
net
j H
g
2
V 
 Actually, since there are 2 to 8 percent nozzle
losses, a velocity coefficient CV is used:
( 0.92  CV  0.98).
net
v
j H
g
2
C
V 

59. Velocity Diagram of
Flow into and out a
Pelton Wheel Bucket

60. Velocity Diagram of Flow out of
a Pelton Wheel Bucket
V2
V2(t)

1800- 
Vj
Vr (2) = k Vr (1)
Vb
Vr (1) = Vj - Vb
k Vr (1) Vb

Vb
Vj
V1 = Vj - Vb
Velocity Diagram of Flow into a
Pelton Wheel Bucket
V2(t) = Vb – (Vj – Vb ) cos (180- )

61. 61
Rotation
Vj (jet velocity) = Vw1
Vb
Vr
u2 = u1
v2
V r, 2
V r,2 = k V r,1
180 -  α2
• Inlet and Exit Velocity Triangular

62. 62
Turbine Selection

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Introduction
In the selection and or design of turbines for a particular
application it is common to rely on model testing. The
results are then scaled up using the following principles
Principles of Similarity Applied To Turbines.
A similar model means:-
Geometrically similar: made from the same drawings but
to a different scale.
Dynamically similar: Operating conditions and equal
efficiencies
Turbine Selection

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Thus in comparing two similar turbines
 All the linear dimensions will be in the same ratio.
 All angles will be the same; the velocity triangles will be
geometrically similar and all velocities will be in the same
ratio.
 The formula suggests that the Pelton turbine is most suitable
for applications with relatively high hydraulic head, due to the
5/4 exponent being greater than unity, and given the
characteristically low specific speed of the Pelton as,
4
5
s H
P
N
N 

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where,
N :is rotational speed of turbine in revelations per minute (rpm),
P: is the break power in kW delivered to the shaft of the turbine, and
H: is net operating head in meter.
(for multi-jet impulse turbines the specific speed is based on the break
power per jet).
 The range of the specific speeds for various turbines is:
 Generally, the turbine manufacturer specifies the specific speed of
its turbine.
Turbine NS
Pelton Wheel 8  Ns  80 High head, low discharge
Radial Flow (Francis) 55  Ns  320 moderate head, moderate discharge
Axial Flow 250  Ns  750 low head, high discharge

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NS is based on the values of N, P, and H used at the design
point. i.e. at maximum efficiency.
 NS is NOT dimensionless and there are different values in
each of the measurement systems.
 The unit of NS is:
The Dimensions of Specific Speed
4
1
2
5
2
1
4
5
2
1
2
L
T
M
L
T
L
T
L
M
T
1










Notes on the Use of the Specific Speed of a Turbine

67. 9/21/2022 67
Practical Consideration

68.  The ideal deflection angle for a Pelton wheel bucket is 
= 180o (i.e.  = 0). From a practical standpoint, the
bucket angle generally chosen is  = 165o (i.e.  = 15o)
or only 2% less than the maximum power, because of
interference of the exiting jet with the incoming jet.
 The number of buckets must be such as to ensure a
smooth interception of the jet for all points on the wheel.

69.  The value of “k” must be as near unity as possible, a
high polish of the bucket surface is essential, and the
the bucket is made as small as possible to reduce
resistance to flow.
 It is possible to have more than one jet operating on a
Pelton wheel and two jets are quite common. Ideally the
power developed is directly proportional to the number
of the jet. However, because the water from one jet
tends to interfere with the water from another,
resulting in reduced efficiency, we now steer away from
multi-jet Pelton wheels.

71. Plan view of a Pelton turbine installation
(Courtesy Voith Siemens Hydro-Power Generation)

72.  The maximum power transmission through a nozzle
occurs when the total pipe loss equals a third of the
available head. The jet diameter Dj which gives the
maximum power efflux from the nozzle is determined
from the nozzle is determined from the following
equation.
where: Dp is the pipe ( penstock ) diameter, and
kn is the nozzle (loss) coefficient.
4
1
p
n
p
j
LD
f
2
1
k
1
D
D












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The Pelton wheel is efficient and reliable when operating under
large heads. To generate a given output power under a smaller
head, the rate of flow through the turbine has to be higher which
requires an increase in the jet diameter.
The number of jets is usually limited to 4 or 6 per wheel.
The increase in jet diameter in turn increases the wheel diameter.
Therefore the machine becomes unduly large, bulky and slow-
running.
In practice, turbines of the reaction type are more
suitable for lower heads
Limitation of a Pelton Turbine

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(a) Spear valve to alter jet area in a Pelton wheel,
(b) Jet deflected from bucket.

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Best
Wishes and
Good Luck

77. 9/21/2022 77
Best Wishes
and Good Luck