Design Analysis of Parts of Pelton Wheel
Turbine
P M V Subbarao
Professor
Mechanical Engineering Department
Selection of right parts and right geometry to
execute Pure Impulse…..
HEPP with Pelton Wheel
Parts of Pelton Turbine
•
•
•
•
•
•
•
The main components of a Pelton turbine are:
(i) water distributor and casing,
(ii) nozzle and deflector with their operating mechanism,
(iii) runner with buckets,
(iv) shaft with bearing,
(v) auxiliary nozzle.
Auxiliary nozzle is used as brake for reducing the speed
during shut down.
• The runner is located above maximum tail water to permit
operation at atmospheric pressure.
Key Parts of Pelton Turbine
Runner with Buckets
• The runner consists of a circular disc with a number (usually more
than 15) of buckets evenly spaced around its periphery.
• Each bucket is divided vertically into two parts by a splitter that has
a sharp edge at the centre and the buckets look like a double
hemispherical cup.
• The striking jet of water is divided into two parts by the splitter.
• A notch made near the edge of the outer rim of each bucket is
carefully sharpened to ensure a loss-free entry of the jet into the
buckets,
• i.e., the path of the jet is not obstructed by the incoming buckets.
Bucket Displacement Diagram
Design of Nozzle is of Prime importance in Pelton Wheel
Nozzle used in 62.5 MW Pelton Wheel
Mechanism of Control of Jet dimensions
The Nozzle and Jet : A Key Step in Design
b
a
djet,VC
Velocity of the jet at VC:
V jet ,VC:ideal  2 gH
d0
V jet ,VC:actual  K v1 2 gH
0.98  Kv1  0.99
Jet carrying a discharge of Q to deliver a power P
P  turbineQgH

 2
Q  d jet ,VC K v1 2 gH
4

To generate a discharge of Q, we need a least jet diameter of
4Q
d jet ,VC 
K v1 2 gH


Diameter of the Jet at the outlet, do

 2
Q  d o K vo 2 gH
4

0.81  K vO  0.83
It is important to find out the VC and outlet jet
diameters/areas
CFD Analysis of Free Jets & Flows In Air
P M V Subbarao
Professor
Mechanical Engineering Department
A Consultancy Project
Sponsored By
BHEL, Bhopal
The set of governing equations solved were primarily the
continuity and the momentum equations.
These basic equations in Cartesian coordinate system for
incompressible flows are given below,

(ui )  0
xi

P  ij
(  ui u j )  

x j
xi x j
 ui u j  2
 ij   

   uk ,k  i , j
 x
 j xi  3
 Turbulent Viscosity
Computational Grid : Nozzle
Contours of Volume Fractions : Nozzle
Air Domain : 0.15m
Contours of Static Pressure : Nozzle
Air Domain : 0.15m
Contours of Velocity Magnitude : Nozzle
Air Domain : 0.15m
Contours of Volume Fractions : Air Domain : 0.3m
Contours of Volume Fractions : Air Domain : 1.0 m
Industrial Correlations for Jet Area
Optimal value of Outlet jet area, ao
ao  As  Bs
2
s is the displacement of spear
b a
A  2ro sin a 
sin b  a 
2
 b  a sin b  sin a sin a
B
2
sin b  a 
Pelton Wheel Distributor - CFD Analysis
• The distributor to the Pelton wheel for the given geometry has
been simulated using Fluent in a 3-d viscous incompressible
flow simulation.
• The set of governing equations solved were primarily the
continuity and the momentum equations.
• The given geometry was meshed using the unstructured
tetrahedral meshes due to geometrical complexity.
• An optimized tetrahedral mesh size of 25 was employed
resulting in a a total of 62583 tetrahedral elements.
CAD Model of Distributor
Pelton Wheel Flow Distributor
Static Pressure Distribution
Distribution of Velocity Magnitude
Mean Diameter of Pelton Runner
Mean diameter or
Pitch circle diameter:
Dwheel
Circumferential
velocity of the wheel,
Uwheel
U wheel  2 gH
U wheel  Kuwheel 2 gH
U wheel  K uwheel 2 gH 
Dwheel 
N wheel
Ns 
4
P
 

gH 
5
Dwheel N wheel
60
60 K uwheel 2 gH
N wheel

 2
Q  d jet ,VC K v1 2 gH
4



 2
P  turbineQgH  gH d jet ,VC K v1 2 gH turbine
4

 60 K u wh eel 2 gH

Dwheel


Ns 






 2

  turbinegH d jet ,VC K v1 2 gH
4




4
gH 5

 2
 60 K uwheel 2 
 

Ns  
  2turbine d jet ,VC K v1 

D
4




wheel


d jet ,VC
 60 K uwheel 

Ns  
turbineK v1 

5
4
Dwheel
2






Experimental values of Wheel diameter to jet diameter
Dwheel /djet,VC
6.5
7.5
10
20
Ns (rpm)
35
32
24
10
0.82
0.86
0.89
0.90
turbine
Ns 
N P
H
5
4
P in hp, H in meters and N in rpm
For maximum efficiency, the ratio should be from 11 to 14.
The highest ratio used in the world is 110 (Kt. Glauraus Power
House in Switzerland).
Specifications of this Pelton wheel are:
Power 3000HP (2.24MW)
Dwheel= 5.36m
Head =1,650 m
N wheel
Ns 
4
Speed: 500 rpm
djet,VC=48.77mm
P

gH 
5

500  3000
4
1650
5
 2.6
Path Lines of Jet
Vj,O
dO
Number of buckets
• The number of buckets for a given runner must be determined so
that no water particle is lost.
• Minimize the risks of detrimental interactions between the out
flowing water particles and the adjacent buckets.
• The runner pitch is determined by the paths of;
• the bucket tip (diameter Dpelton), the Wheel diameter (DWheel).
• and the relative paths of the water particles stemming from the
upper and lower generators of the jet.
• The bucket pitch must be selected so that no particle stemming
from the lower generator of the jet can escape the runner without
encountering any bucket.
Bucket Duty Cycle
Reference Position
Zones of Bucket Duty Cycle
•
•
•
•
•
•
i) Approach of the tip to the jet (θj < −40◦).
ii) Initial feeding process : (θj = −40◦...−10◦).
iii) Entire separation of the jet (θj = −10◦...0◦)
iv) Last stage of inflow (θj = 0◦...15◦)
v) Last stage of outflow (θj = 15◦...50◦).
vi) Series of droplets (θj = −50◦...∞).
Start of Jet Bucket Interactions
Sequence of Jet Bucket Interactions
Dq50
Dq150
Dq250
Dq350
Dq450
Dq550
Minimum Number of Buckets
The axis of the jet falls on Pitch Circle

w
2j
1D
Dj,O,
Vj,O
1A
1B
1C
Minimum Number of Buckets

w
q
2j  qy
y
dj,O, Vj,O
1B
1E
1C
Minimum Number of Buckets
tj : Time taken
bye the jet to
travel lj

tb: Time taken by
first bucket to
travel y
w
q
dO, Vj,O
lj
y
• tj = lj/Vjet,O
• tb = y/w
l j  D pelton sin 
U wheel
w
Rwheel

RP
dO,
Vj,O
RW w
q
lj
y
lj
For better working tj < tb
D pelton sin 
V j ,O

yRwheel
U wheel
V j ,O
y
y

w
U wheel D pelton sin 
RwheelV j ,O
The minimum allowable value of y
ku ,wheel
U wheel Dpelton
y
sin  
V j ,O Rwheel
kvO

 
21 
 sin 
 Rwheel 
cos  
Rwheel 
cos  
1
d jet ,O
Dwheel
cos  
2
1
Dwheel
2
R pelton
Rwheel 
d jet ,O

d jet ,VC
 60 K uwheel 2 

Ns  
 turbineK v1 
Dwheel


 4 

2
Rwheel  
1

d jet ,O
Dwheel
cos  
2
1
Dwheel
d jet ,O
RP
dO,
Vj,O
RW w
q
lj
y
Maximum allowable angle between two successive buckets
q  2 y
Minimum number of buckets
z
360
q
Dr Taygun has suggested an empirical relation for z
Dwheel
z  0.5
 15
d jet ,VC
Bucket Power Distribution
P(qj)
Total
2
1
4
3
5
Bucket Energy Distribution
E
k
Bucket Energy
h 
 k
Water energyinterscpet d by the bucket mwater gH
Ej,k
Geometric Details of Bucket
The hydraulic efficiency
depends more on the main
bucket dimensions (length
(A), width (B) and depth
(C)).
The shape of the outer part
of its rim or on the lateral
surface curvature also has
marginal effect on
hydraulic efficiency.
Shape variations of Buckets
Design 1
Design 2
Empirical Geometry of Bucket Shape
III
II
C
IV
V
I
DW
S
2bi
A
be
B
Empirical Relations for Bucket Geometry
•
•
•


•
•
•
•
•
A = 2.8 djet,VC to 3.2 djet,VC
B = 2.3 djet,VC to 2.8 djet,VC
C= 0.6 djet,VC to 0.9 djet,VC
bi = 50 to 80
be is varied from section I to section V
I: 300 to 460
II: 200 to 300
III: 100 to 200
IV: 50 to 160
V: 00 to 50