Last Rev.: 19 JUN 08
Pelton Wheel: MIME 3470
Page 1
Grading Sheet
~~~~~~~~~~~~~~
MIME 3470—Thermal Science Laboratory
~~~~~~~~~~~~~~
Laboratory №. 11
PELTON WHEEL
PERFORMANCE CHARACTERISTICS
Students’ Names  Section №
POINTS
PRESENTATION—Applicable to Both MS Word and Mathcad Sections
GENERAL APPEARANCE, ORGANIZATION, ENGLISH, & GRAMMAR
5
ORDERED DATA, CALCULATIONS & RESULTS—MATHCAD
30
25
CALCULATIONS
COMBINED PLOT
DISCUSSION OF RESULTS
WHAT ARE TWO FEATURES OBSERVED FROM COMBINED PLOT?
WHY IS THE BUCKET EXIT ANGLE SMALL AND NONZERO?
WHAT IS PURPOSE OF CUTOUT AT BUCKET PERIPHERY?
WHAT IS THE MAXIMUM UTILIZATION FACTOR WHEN 2 = 0?
WHY MUST A TURBINE W/ NONZERO HAVE A ROTOR ENCLOSED?
WHY DOES THE RELATIVE VELOCITY REMAIN UNCHANGED AS
THE FLOW PASSES THROUGH A PELTON BUCKET?
CONCLUSIONS
ORIGINAL DATASHEET
TOTAL
COMMENTS
d
GRADER—
5
5
5
5
5
5
5
5
100
SCORE
TOTAL
Last Rev.: 19 JUN 08
Pelton Wheel: MIME 3470
MIME 3470—Thermal Science Laboratory
~~~~~~~~~~~~~~
Laboratory №. 11
PELTON WHEEL
PERFORMANCE CHARACTERISTICS
~~~~~~~~~~~~~~
LAB PARTNERS: NAME
NAME
NAME
SECTION
№
EXPERIMENT TIME/DATE:
NAME
NAME
NAME
TIME, DATE
~~~~~~~~~~~~~~
OBJECTIVE—of this experiment is to plot the Pelton wheel actual
and theoretical horsepowers vs. U/V1 (explained below) and to
demonstrate that maximum power occurs when U/V1  0.5.
INTRODUCTION
A water turbine1 is a rotary engine2 that takes energy from moving
water. The Pelton wheel is such a device.
The earliest water turbines were water wheels which have been
used for thousands of years for industrial power. Their main shortcoming is size, which limits the flow rate and head that can be used.
The migration from water wheels to modern, more efficient turbines took during the 19th century during the Industrial Revolution.
Efficiency improvements of water turbines allowed them to compete
with steam engines (wherever water was available). These turbines
(using scientific principles, new materials, and, new manufacturing
methods) were widely used for industrial power prior to electrical
grids. Now they are mostly used for electric power generation.
The main difference between early water turbines and water
wheels is a swirl component of the water which passes energy to a
spinning rotor. Swirl is the tangential velocity component induced
by a curved impeller. This additional component of motion allowed
the turbine to be smaller than a water wheel of the same power. They
could process more water by spinning faster and could harness much
greater heads. Later, impulse turbines such as the Pelton wheel
were developed which did not use swirl.
THEORY
Euler Turbine Equation
ALL turbines have a runner or rotor which holds the turbine vanes
or blades. In the case of a water wheel, the vanes are simple
paddles. This runner and attached vanes rotate as flowing water is
directed onto the vanes. Since the runner is spinning, the imparted
water force acting through a distance generates work. In this way,
energy is transferred from the water flow to the turbine.
The basic design relationship for all turbomachines is very simple
and is only a form of Newton’s Laws of Motion applied to fluid
traversing a rotor. Imagine water impinging on a turbine runner at
Point 1 at some arbitrary angle and at rotor radius r1. Further, the
fluid exits the runner at Point 2 having some other arbitrary angle
and at rotor radius r2. Also, the flow is assumed to be an
unchanging steady-state at every point in the system. The fluid
velocity vector at any point can be resolved into three mutually
perpendicular components:
Va – an axial component directed parallel to the axis of rotation,
Vm – a radial component directed radially outward from the axis
of rotation, and
Vu – a tangential component directed parallel to the tangential
velocity of the rotor, U.
Page 2
The change in magnitude in axial and radial components
causes forces must be carried by the bearings. These forces,
however, do not affect the angular rotation of the rotor except
through bearing friction.
However, the change in magnitude and of radius of the
tangential velocity components corresponds to a change in angular
momentum of the fluid and by Newton’s Laws of Motion is equal
to the summation of all the applied forces on the rotor; i.e., the net
torque on the rotor, . In general terms, this is as follows. If a mass
of fluid M1 enters the rotor at radius r1, with tangential component
of absolute velocity Vu1 during time t; and, if during the same time
t a mass M2 leaves the rotor at radius r2 with tangential component
of absolute velocity Vu 2 ; then

M1
M
r1Vu1  2 r2Vu 2
g ct
gct
For a unit mass steady flow, M1 t  M 2 t  m  1 this becomes
  r1Vu1 g c  r2Vu 2 g c . For a steady rotor angular velocity of
, the rate of energy transferred or utilized Eutil = . Further, the
linear velocity of the rotor is U = r. Combining all this yields
1
(1)
Eutil 
U1Vu1  U 2Vu 2
gc
This is one form of the Euler turbine equation or simply the Euler
equation.3


Degree of Reaction
The relative proportions of energy transfer obtained by change
of both static and dynamic pressures are important factors with
respect to classifying turbomachines, as for a given class of
machine this proportion inevitably leads to a particular type of
design with certain inherent characteristics. The parameter used to
describe this relation is the degree of reaction or more simply the
reaction, R, which is defined as the ratio of the energy transfer by
means of or resulting in a change of static pressure in the rotor to
the total energy transfer to (utilized by) the rotor.
1
U12  U 22  Vr2  Vr2
2
1
2gc
R
(2)
1
U1Vu1  U 2Vu 2
g
c





Eutil
where,
Eutil  Total energy transferred to the rotor
U  linear velocity of the rotor at radius r, = r
V  absolute velocity
Vu  tangential component of absolute velocity
Vr  velocity relative to the moving rotor
R, can be negative, zero, and values greater than unity.
Reaction Turbines
Water turbines are either reaction turbines or impulse turbines. All
common water turbines until the late 19th century were reaction
turbines. Reaction turbines are acted on by water, which changes
static pressure as it moves through the turbine and gives up its
energy. The flow between the rotating vanes of a reaction turbine can
be envisioned as flow through rotating nozzles. The pressures within
the moving nozzle produce work. The flow must be encased to
contain the water pressure (or suction), in order that the fluid cannot
1
turbine: word coined by the French engineer Claude Bourdin in the early
19th century and is derived from the Latin word for “whirling” or a “vortex”.
2
engine: a mechanical device producing some form of output from a given
input.
3
This should not be confused with the Euler flow equation which is simply
the Navier-Stokes equation without viscosity effects.
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Pelton Wheel: MIME 3470
expand freely in all directions. Alternately, the vanes must be fully
submerged in the water flow as with a ship’s propeller.
Most water turbines in use are reaction turbines—even a
simple spinning lawn sprinkler. They are used in low and medium
head applications. Two popular types of reaction turbines are the
Francis and the Kaplan turbines.
In 1826, Benoit Fourneyron developed a high efficiency (80%)
outward flow water turbine. Water was directed tangentially through
the turbine runner causing it to spin. Jean-Victor Poncelet designed
an inward-flow turbine in about 1820 that used the same principles.
S. B. Howd obtained a U.S. patent in 1838 for a similar design.
In 1848, James B. Francis improved on the inward-flow design to
create a turbine with 90% efficiency. It is also called a radial-flow
turbine He applied scientific principles and testing methods to
produce the most efficient turbine design ever. More importantly, his
mathematical and graphical calculation methods improved the state
of the art of turbine design and engineering. His analytical methods
allowed confident design of high efficiency turbines to exactly
match a site’s flow conditions.
Francis turbines are the most common water turbine in use today.
They operate in a head range of ten meters to several hundred
meters and are primarily used for electrical power production.
Figure 1—Francis Turbine: (a) Cross section,
(b) Runner used in Grand Coulee Dam
The Kaplan turbine is an inward-flow, propeller-type water
turbine that has adjustable blades developed in 1913 by the Austrian
professor Viktor Kaplan. It was an evolution of the Francis turbine.
Figure 2—Kaplan Propeller Turbine
Its invention allowed efficient power production in low head
applications that was not possible with Francis turbines. Kaplan
turbines are now widely used throughout the world in high-flow,
low-head power production. Their efficiencies are typically over
90%, but may be lower in very low head applications. Again, most
water turbines in use are reaction turbines. They are used in low
and medium head applications.
Impulse Turbines
Zero reaction (R = 0) is an important value and characterizes a
particular design of many types of turbomachine. The Knight and
Pelton bucket wheels are examples of zero reaction turbines.
In 1866, California millwright Samuel Knight invented a
machine that worked from a different concept. Inspired by the high
pressure jet systems used in hydraulic mining in the gold fields,
Knight developed a wheel where the vanes were bucket-like which
captured the energy of a free jet, which had converted a high head
(hundreds of vertical feet in a pipe or penstock) of water to kinetic
Page 3
energy. This is called an impulse or tangential turbine. The water's
velocity, roughly twice the velocity of the bucket periphery, does a
U-turn in the bucket and drops out of the runner at low velocity.
In 1879, Lester Pelton4, experimenting
with a Knight Wheel, developed a doublesemicylindrical bucket design, which
exhausted the water to the side, eliminating
some energy loss of the Knight wheel which
exhausted some water back against the center
of the wheel. In about 1895, William Doble
improved on Pelton's half-cylindrical bucket
form with an elliptical bucket that included a
cutout in it to allow the jet a cleaner bucket
Lester Allen Pelton
exit by centrifugal force. This is the modern
form of the Pelton turbine which today achieves up to 92% efficiency.
Pelton had been quite an effective promoter of his design and
although Doble took over the Pelton company he did not change the
name because it had brand name recognition.
So, if R = 0, there is no change of static pressure in the rotor
and such a turbine is an impulse-type. In these machines, prior to
hitting the buckets, the water’s head pressure (potential energy) is
entirely converted to kinetic energy by a nozzle and focused on the
turbine (see Figure 3). As the nozzle is stationary, it performs no
work even though force is transferred to it as it accelerates the
flow. As no pressure change occurs at the turbine blades the
turbine doesn't require an enclosing housing.
Utilization Factor
The quality of interest in a turbine is the adiabatic efficiency which
is usually taken to be equal to the overall efficiency, since mechanical
efficiency is nearly unity. However, the adiabatic efficiency is also
the product of two factors, the first factor being referred to as the
utilization factor or diagram efficiency is the ratio of the ideal
work output to the energy available for conversion into work. The
value of the utilization factor may be found from the ideal velocity
4
LESTER ALLAN PELTON: FATHER OF HYDROELECTRIC POWER
September 5, 1829 – March 14, 1908
Lester Pelton was born in Vermilion Township (Erie County), Ohio. In
the spring of 1850, Lester and some other youths headed for gold country in
California. Arriving there Pelton left his friends and went to Sacramento
where he peddled fish to miners. After hearing that digging gold in the Sierra
Nevada Mountains was more profitable he moved some ninety miles north to
Camptonville along the Yuba River in 1860. All types of mining were going
on there; placer, hard rock, and hydrologic. Although Lester was not terribly
interested in mining he was an avid reader and he enjoyed watching the mining
efforts. A very introspective person, he was also a skilled tinsmith, carpenter,
and millwright. At the time steam engines powered most of the mining works.
According to a 1939 article by W. F. Durand of Stanford University in
Mechanical Engineering: “Pelton's invention started from an accidental
observation, some time in the 1870s. Pelton was watching a spinning water
turbine when the key holding its wheel onto its shaft slipped, causing it to
become misaligned. Instead of the jet hitting the cups in their middle, the
slippage made it hit near the edge; rather than the water flow being
stopped, it was now deflected into a half-circle, coming out again with
reversed direction. Surprisingly, the turbine now moved faster. That was
Pelton's great discovery. In other turbines the jet hit the middle of the cup
and the splash of the impacting water wasted energy.”
Pelton’s wheel was first used at the Mayflower Mine in Nevada City,
California in 1878. By 1879 he had tested a prototype Pelton wheel at the
University of California. In 1887 a miner attached Pelton's wheel to a dynamo
and produced the first hydroelectric power in the Sierra Nevada Mountains.
With Pelton’s wheel, low-cost hydroelectric power could replace expensive
steam engines in mining operations in the western states, where streams
rarely flowed at high enough volumes to turn traditional water wheels.
A patent was granted in 1889 to Pelton. To keep up with tremendous
demand, Pelton and a San Francisco machine shop owner organized the
Pelton Water Wheel Company. Today, Pelton’s wheel still generates
electricity in small hydroelectric power plants in the western United States.
Last Rev.: 19 JUN 08
Pelton Wheel: MIME 3470
diagrams and the energy equations. Under ideal conditions, it
should be possible to use all of the available kinetic energy of the
liquid at the turbine inlet and also the energy obtained in the rotor
due to pressure drop (i.e., reaction effort). Since the former
quantity is V12 / 2gc , while the latter quantity is
U
2
1



 U 22 / 2 g c  Vr2  Vr2 / 2 g c
2
1
the ideal energy available for conversion into work in the turbine is:
1
Eavail 
V12  U12  U 22  Vr2  Vr2
2
1
2gc
On the other hand, the work output (energy utilized) from the
system is given by the Euler turbine equation (see Equation 1).
The ideal utilization factor, , is the ratio of Eutil to Eavail, i.e.,
 

Eutil

Eavail




1
U1Vu1  U 2Vu 2
gc
 



U12
 U 22
 
Vr2
2
1 = 0


2 U1Vu1  U 2Vu 2
The theoretical maximum value of utilization factor should
occur when

d
U
U 1
 0  21  cos  2 1  2  or

V
V
2
U 
1
1

d  
V
 1

U 1
d
U
(5)
 0  21  cos  2 1  2  or

V1 
V1 2
U 

d  
 V1 
i.e., when the linear velocity of the bucket is half that of the
absolute velocity of the impinging jet.

 Vr2
1


V12
2U V1  U 1  cos  2 
V12
SPEAR
SIDE OF BUCKET
PROCEDURE
Given the apparatus of Figure 4, do the following:
1. In Figure 4, is a box labeled Pump Motor Controller. With this,
one dials in a nominal percent of maximum pump motor rpm.
For 60%, 50%, and 40% settings, perform the following steps.
2. For each of these settings determine the following:
a. Determine the flow rate using the Volumetric Measuring
Tank and a stopwatch. The lab instructor will demonstrate
how to fill the tank.
b. Measure the pump discharge pressure and note the pump
discharge inner diameter. From these, one can determine the
absolute velocity of the jet from the Pelton nozzle, V1, from
Bernoulli’s equation.
c. Note the diameters of the Pelton wheel (to the center of the
cups) and of the Prony brake.
d. The Prony brake6, shown in Figure 4 and at the right, is used
to both apply a torque to the Pelton wheel and to measure
the torque generated. For a Prony brake
radius of Rbrake , that torque is
  Rbrake F1  F2
The forces F1 and F2 are determined from
Prony Brake
spring-loaded scales (see Figure 4).
e. For a given flow rate, one wants to first
determine the Pelton wheel rpm with no Prony brake load,
rpmmax, and that Prony brake load (F1 and F2) that just stops
the Pelton wheel, rpm = 0. Rpm is measured using a laser
tachometer.
The Pelton-wheel bucket velocity, U, with no Prony brake
load can be considered a good approximation to the actual jet
velocity, V1. Since the torque, , is zero for this condition,
horsepower is also zero (HP = ).
  2UVr 1  cos  2 
1
V12
2

U U  
 21  cos  2     
 V1  V1  


Also known as zero-angle type.
TAIL RACE
Figure 3—Pelton Wheel Impulse Turbine
(4)
6
5
VARIABLE
AREAS
NOZZLE
CUTOUT
where 2 is described in Figure 3.
Thus the utilization factor for a Pelton wheel turbine becomes
 
V2

Vu 2  U  Vr2 cos  2

U
Vr2
(3)
change in the velocity’s direction. This resulting change in
momentum (impulse) causes a force on the turbine buckets. The
denominator of Equation 3 representing energy available then
becomes simply V12.
The Vu terms in the numerator of Equation 3 can be simplified
by referring to the vector diagram of Figure 3.
Vu1  V1  U  Vr1
2U U  Vr1  U  Vr2 cos  2
Vr1
U
2
Pelton Wheel
The Pelton wheel is composed of a nozzle which converts the
whole available head to kinetic energy and a rotor made up of a
series of double hemispherical buckets fastened on the periphery
of the rotor. The rotor is not enclosed, and the water leaving the
buckets goes immediately to the tailrace.
The Pelton wheel falls in a large class of these machines known
as the axial-flow type5 where the nozzle angle relative to the runner is
zero. However, the bucket cannot have a 180° camber angle, since
the water must have a finite radial velocity component away from
the wheel in order to avoid interference.
The Pelton wheel shown in Figure 3 is a pure impulse (R = 0)
turbine. It is used in very high head installations and develops efficiencies very close to the Francis and Kaplan reaction turbines.
The utilization factor described in Equation 3 can be further
simplified as follows: For these machines there is no change of
rotor radius, so U1 = U2. Further, since the energy transfer is entirely
at atmospheric pressure (see Figure 3) the absolute flow velocity
remains unchanged; i.e., an impulse machine of the axial-flow type
has Vr1  Vr2 . Thus the energy transfer is wholly derived from a

WATER
JET
V1
1
V12  U12  U 22  Vr2  Vr2
2
1
2gc
V12
Page 4
prony brake: a simple device invented by Gaspard de Prony to measure the
torque produced by an engine. The term brake horsepower is one measurement of torque obviously derived from this method of measurement.
Last Rev.: 19 JUN 08
Pelton Wheel: MIME 3470
With the wheel just stopped with the Prony brake, the
horsepower should also be zero.
There is space on the data sheet for 10 rpms for each flow
condition. In the procedure just above, the first and last rpms
are defined. Now, divide the maximum rpm by 10 to determine
the rpm interval such that the remaining 8 data points can be
determined. Measure the Prony brake loads for each rpm.
Pump and Motor Characteristics
Impeller OD:
5”
№ of Impeller Blades: 6
Type of Impeller:
Open Type of Blades:
Backward Curving
Torque Arm Length: 5”
Motor Speed Range: 0-3000 rpm
SpringLoaded
Scale 
Page 5
CALCULATIONS
A Mathcad is supplied below. Use
the variable names already supplied
in the object. At the bottom of this
object is in the required format the
student is to supply a summary of
the 60%-rpm condition. This along
with the required plot should be
more than sufficient to evaluate the
student’s work. One should be able
to
put all their Mathcad
calculations on one sheet as shown
in Figure 5.
Figure 5—Example OneThe required plot is as shown in
Page Calculation Sheet
Figure 6. The data used to generate
this plot is questionable so the actual shape of the curves is subject
to change.
Prony Brake 
Pump
Pump
Motor
Motor
Controller
Controller


Pelton Wheel 
Buckets
Motor

Pump
(see below)
Volumetric
Measuring
Tank
 Torque Arm
Sump Tank
Weight Hanger 
Figure 6— Sample Plot of Calculations
Note: The data used for this plot is questionable
Also answer the questions found in the Discussion of Results and
supply a Conclusion(s).
Figure 4—Pump Performance Test Rig and Pelton Wheel
REFERENCES
1. Principles of Turbomachinery, D.G. Shepherd, MacMillian
Publishing Co., Inc, New York, 1956.
2. Wikipedia,
http://en.wikipedia.org/wiki/Water_turbine
http://en.wikipedia.org/wiki/Pelton_wheel
http://en.wikipedia.org/wiki/Francis_turbine
http://en.wikipedia.org/wiki/Kaplan_turbine
3. Engineering Fluid Mechanics, J.A. Roberson and C.T. Crowe,
John Wiley & Sons, Inc., 6th Edition, 1997.
4. Turbines, J.B. Calvert, University of Denver
http://mysite.du.edu/~jcalvert/tech/fluids/turbine.htm
5. National Inventors Hall of Fame
http://www.invent.org/hall_of_fame/293.html
6. An Introduction to Energy Conversion: Turbo Machinery, Vol. 3,
V.Kadambi and Manohar Prasad, Wiley Eastern, New Delhi, 1977
Google Books Result
Last Rev.: 19 JUN 08
Pelton Wheel: MIME 3470
Page 6
ORDERED DATA, CALCULATIONS, and RESULTS (Mathcad object is reduced to 80%)
i  1  3
DATA: TRIAL INDEX:
RUN INDEX:
j  1  10
ID  1.5  in
PUMP DISCHARGE ID:
PELTON WHEEL RADIUS: R  5.25  in
TRIAL 1 (i = 1)
NOMINAL 60% OF MAX PUMP RPM
F1
1j

F2
1j

NP
1j

Vol 
1
t 
1
H 
1
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
PRONY BRAKE RADIUS: RBrake  3  in
WATER DENSITY:   62.24  lb  ft
TRIAL 2 (i= 2)
TRIAL 3 (i = 3)
|
NOMINAL 50% OF MAX PUMP RPM
NOMINAL 40% OF MAX PUMP RPM
|
F1
 F2
 NP

F
 F2
 NP

| 13  j
2j
2j
2j
3j
3j
|
Vol 
|
3
Vol 
2
|
|
t 
t 
2
3
|
|
H 
H 
|
2
3
|
|
|
|
|
|
|
3
SUMMARY OF TRIAL 1 (60%): m_dotAct  kg
1
V 1.Act  ft
1
T Act
1j
ft  lbf

V 1.Th  ft
1
HP Act
1j
ft  hp

U

1j
ft
Last Rev.: 19 JUN 08
Pelton Wheel: MIME 3470
DISCUSSION OF RESULTS
What two features can be observed from the plot of the results?
Answer:

Why are the buckets designed so that 2 (see Figure 3) is a small,
nonzero angle?
Answer:
What is the purpose of the cutout or notch at the periphery of the
Pelton wheel buckets?
Answer:
What is the maximum ideal utilization factor for a Pelton wheel
when 2 = 0?
Answer:
Why must a turbine with nonzero degree of reaction have a rotor
enclosed?
Answer:
Why does the relative velocity remain unchanged as the flow
passes through a Pelton bucket?
Answer:
CONCLUSIONS
Page 7
Last Rev.: 19 JUN 08
Pelton Wheel: MIME 3470
Page 8
APPENDIX—DATA SHEET
Time/Date:
___________________
Lab Partners:
_______________________
_______________________
_______________________
_______________________
_______________________
_______________________
Particulars of the Apparatus: Pelton Wheel Radius, R:
_________ (
) Prony Brake Radius, RBrake: _________ (
Pump Discharge Diam, ID: _____1.5 in___
Trial 1
(i= 1)
Nominal
60% of
Max
Motor
RPM
Volume, Vol
(
Trial 2
(i= 2)
Nominal
50% of
Max
Motor
RPM
)
Pump
Flow Rate
Time, t
Pump Discharge Gage Pressure, H
(
)
(
)
Prony Brake
Forces
Run, j
Pelton Wheel rpm
F2
)
(
)
Time, t
(
)
(
)
Pump Discharge Gage Pressure, H
Prony Brake
Forces
F1
(
Volume, Vol
Pump
Flow Rate
(
Run, j
)
F1
(
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
Trial 3
(i= 3)
Nominal
40% of
Max
Motor
RPM
(
)
Time, t
(
)
(
)
Pump Discharge Gage Pressure, H
Prony Brake
Forces
Run, j
F1
(
1
2
3
4
5
6
7
8
9
10
Pelton Wheel rpm
F2
)
(
Volume, Vol
Pump
Flow Rate
(
)
Pelton Wheel rpm
F2
)
)
)